ch15 z5e aq. equil

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Chapter 15

Applying equilibrium pp

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15.1 The Common Ion Effect pp

When the salt with the anion of a weak acid is added to that acid,

It reverses the dissociation of the acid. Lowers the % dissociation of the acid. HAc

H1+ + Ac1- (add NaAc)

Same for salts with cation of weak base. H2O + NH3 NH4

1+ + OH1- (add NH4Cl) The calculations are the same as last

chapter.

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15.2 Buffered solutions pp A solution that resists a change in pH. Consists of either a weak acid and its salt

or a weak base and its salt. We can make a buffer of any pH by

varying the concentrations of these solutions.

Same calculations as before. Calculate pH of a solution of .50 M HAc

& .25 M NaAc (Ka = 1.8 x 10-5). Steps follow.

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pH of .50 M HAc & .25 M NaAc pp

Write major species: HAc, Na1+, Ac1-, H2O R HAc H1+ + Ac1- I 0.50 0 0.25

C - x + x + x E 0.50 - x x 0.25 + x

Assume x <<, since Ka = 1.8 x 10-5 (x)(.25)/(.50) = 1.8 x 10-5 [H1+] = ??? [H1+] = 3.6 x 10-5 pH = ??? . . . pH = 4.44

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Adding a strong acid or base pp

Do the stoichiometry first (use moles). Think about the chemistry, write major

species & check to see an =m reaction will occur among them.

A strong base will grab protons from the weak acid reducing [HA]0

A strong acid will add its proton to the anion of the salt reducing [A-]0

Then do the =m problem in molarity.

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Another way to think of it pp

Write major species & look for a strong something & a weak something.

The stoich will be between them (SA + WB or SB + WA) in moles.

E.g., If major species are HAc, Ac1-, H1+

(from HCl) & H2O, the stoich is . . . H1+ + Ac1- HAc SA

WB WA Then do the =m problem in molarity.

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Adding a strong acid or base to buffered soln pp

What is pH of 1.0 L of 0.50 M HAc + 0.25 M NaAc after add 0.010 mol solid NaOH?

Write major species . . . HAc Na1+ Ac1- OH1- H2O Look for a stoich reaction in moles . . . HAc + OH1- HOH + Ac1- i 0.50 0.01 0.25 mol

f 0.49 0 0.26

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HAc + OH1- HOH + Ac1- i 0.50 0.01 0.25 mol f 0.49 0 0.26

Now change back to M (had 1 L) & do =m R HAc H1+ + Ac1- I 0.49 0 0.26 C

- x + x + x E 0.49 - x x 0.26 + x

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R HAc H1+ + Ac1- I 0.49 0 0.26 C - x + x + x E 0.49 - x x 0.26 + x

Assuming x << than 0.49 or 0.26(x)(0.26)/(0.49) = 1.8 x 10-5

X = [H1+] = ??? 3.4 x 10-5 pH = ??? 4.47 (compare to 4.44 before adding OH1- )

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What if there was no buffer? pp

What is pH of 1.0 L of 0.50 M HAc after adding 0.010 mol of solid NaOH? (pH of HAc before adding OH1- is 2.52)

Write major species . . . HAc Na1+ OH1- H2O Look for a stoich reaction in moles . . . HAc + OH1- HOH + Ac1- i 0.50 0.01 0 mol

f 0.49 0 0.01

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What if there was no buffer? pp HAc + OH1- HOH + Ac1- i

0.50 0.01 0 mol f0.49 0 0.01

Now to M & do =m; Ka = 1.8 x 10-5 R HAc H1+ + Ac1- I 0.49 0 0.01 C

- x + x + x E 0.49 - x x 0.01 + x

pH = ??? pH = 3.05, compare to 2.52 for HAc alone.

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General equation Ka = [H+] [A-]

[HA] so [H+] = Ka [HA]

[A-] The [H+] depends on the ratio [HA]/[A-] taking the negative log of both sides pH = -log(Ka [HA]/[A-]) pH = -log(Ka) + -log([HA]/[A-]) pH = pKa + log([A-]/[HA])

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This is called the Henderson-Hasselbach equation

pH = pKa + log([A-]/[HA])

pH = pKa + log([base]/[acid]) Only use this (instead of RICE) when

there is a BUFFER! It’s a nice short cut for buffers, but

students tend to use it incorrectly for other problems, so be careful.

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Sample Problem pp

pH = pKa + log([base]/[acid]) Use H/H to calculate the pH of 0.75 M

lactic acid (HC3H5O3) and 0.25 M sodium

lactate (Ka = 1.4 x 10-4) . . .

pKa = ???

pKa = 3.85 pH = 3.85 + log (0.25/0.75) = ??? pH = 3.37 Can also do by the RICE method.

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Another Sample Problem pp pH = pKa + log([base]/[acid])

Calculate the pH of 0.25 M NH3 and 0.40 M

NH4Cl (Kb = 1.8 x 10-5) . . .

Whoa! We have Kb. Are we stuck?

No! Ka x Kb = Kw = 1.0 x 10-14 Your answer is . . . pH = 9.05 Calculation . . . pH = (14 - pKb) + log(.25/.40) = 9.05 Can get the same using RICE, but will get pOH

and must subtract from 14 to get pH.

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Hints for doing the online HW pp Most of the problems will have a part to them Most of the problems will have a part to them

that you can solve using the that you can solve using the H-H equationH-H equation.. If there is a neutralization reaction then do the If there is a neutralization reaction then do the

stoichiometrystoichiometry first to determine how the first to determine how the mmolesmmoles of HA and A of HA and A1-1- have changed. Then, have changed. Then, convertconvert to [ ] and use the to [ ] and use the H-H equationH-H equation..

Remember that Remember that ifif you have equal you have equal equilibriumequilibrium amounts of HA and Aamounts of HA and A1-1- then then pkpkaa = pH = pH. .

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Hints for doing the online HW pp Acids & Bases dissociate Acids & Bases dissociate stepwisestepwise, so even if , so even if

you don’t know the acid (e.g, benzoic acid) you don’t know the acid (e.g, benzoic acid) you know that it loses only you know that it loses only oneone hydrogen hydrogen (unless you’re told it is diprotic, triprotic, etc.)(unless you’re told it is diprotic, triprotic, etc.)

The term “The term “ionization constantionization constant,” as used in the ,” as used in the online HW, is a online HW, is a genericgeneric term for k term for kaa or k or kbb

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Let’s Prove They’re buffers pp

What is the pH if .10 mol of gaseous HCl is added to 1.0 L of 0.25 M NH3 &

0.40 M NH4Cl (Kb = 1.8 x 10-5)? (Why gaseous? Why 1.0 L?)

Gaseous or solid means total volume doesn’t change.

1.0 L means mols = molarity Be careful! Convert mols back to

molarity since volume might change!! See summary, p. 692

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What is the pH if .10 mol of gaseous HCl is added to 1.0 L of 0.25 M NH3 &

0.40 M NH4Cl (Kb = 1.8 x 10-5)? Steps

follow. Do the stoich to get . . . H1+ + NH3 NH4

1+ i .10 .25 .40

f 0 .15 .50

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Let’s Prove They’re buffers pp pH if .10 mol HCl(g) added to 1.0 L of 0.25

M NH3 & 0.40 M NH4Cl Kb = 1.8 x 10-5

After stoich, 0.15 M NH3 & 0.50 M NH41+

Do the Equilibrium to get . . . R NH3 + H2O NH4

1+ + OH1- I .15 .50 0

C - x +x +xE .15 - x .50 + x x

Etc. or Use Henderson-Hasselbach! . . .

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pH if .10 mol HCl(g) added to 1.0 L of 0.25 M NH3 & 0.40 M NH4Cl Kb = 1.8 x 10-5

Using H-H pH = -log(10-14/1.8 x 10-5) + log(.15/.50) pH = 8.73 (Compare to original of 9.05)

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What would the pH be if 0.050 mol of solid NaOH is added to 1.0 L of 0.75 M lactic acid HC3H5O3) & 0.25 M sodium

lactate (Ka = 1.4 x 10-4) Do Stoich, then H-H equation to get ??? pH = 3.48, compare to original of 3.38. Be careful! Convert mols back to

molarity since volume might change!!

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15.3 Buffer capacity The pH of a buffered solution is

determined by the ratio [A-]/[HA]. As long as this ratio doesn’t change

much the pH won’t change much. The more concentrated these two are the

more H+ and OH- the solution will be able to absorb.

Larger concentrations means bigger buffer capacity.

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Buffer Capacity - Use “RIF” & H-H method for these

Calculate pH Calculate pH changechange when 0.010 mol HCl when 0.010 mol HCl(g)(g) is is

added to 1.0L solutionadded to 1.0L solutionss of HAc & NaAc . . . of HAc & NaAc . . . (Ka = 1.8x10(Ka = 1.8x10-5-5 & pH & pH withoutwithout HCl = 4.74) HCl = 4.74)

Write the reaction (SA + WB), which is . . .Write the reaction (SA + WB), which is . . . HH1+1+ + Ac + Ac1-1- HAc HAc (HAc not always on reactant side)(HAc not always on reactant side)

5.00 5.00 MM HAc & 5.00 HAc & 5.00 MM NaAc. pH = ??? NaAc. pH = ??? pH still 4.74 (insignificant change)pH still 4.74 (insignificant change) 0.050 0.050 MM HAc & 0.050 HAc & 0.050 MM NaAc. pH = ??? NaAc. pH = ??? pH = 4.56 so pH = 4.56 so changechange is only is only -0.18-0.18

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Buffer capacity The best buffers have a ratio

[A-]/[HA] = 1 Memorize this rule of thumb! This is most resistant to change True when [A-] = [HA] Make pH = pKa (since log1 = 0)

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Buffer capacity Which acid (& its sodium salt) below will

give the best buffer at pH = 4.30?Chloracetic acid (Ka = 1.35 x 10-3)Propanoic acid (Ka = 1.3 x 10-5)Benzoic acid (Ka = 6.4 x 10-5)Hypochlorous acid (Ka = 3.5 x 10-8) . Hint: don’t find actual [ ]s of HA and A-; just their ratio. Answer ??? . . .

Benzoic Acid (pKa = 4.19, closest to 4.30)

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15.4 Titrations & pH Curves Millimole (mmol) = 1/1000 molMillimole (mmol) = 1/1000 mol Molarity = mmol/mL = mol/L Molarity = mmol/mL = mol/L Makes calculations easier because we Makes calculations easier because we

will rarely add Liters of solution.will rarely add Liters of solution. Adding a solution of known Adding a solution of known

concentration until the substance being concentration until the substance being tested is consumed.tested is consumed.

This is called the This is called the equivalence pointequivalence point.. Graph of pH vs. mL is a titration curve.Graph of pH vs. mL is a titration curve.

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Strong acid with Strong Base pp Do the Do the stoichiometrystoichiometry.. There is There is no equilibriumno equilibrium . . They both dissociate completely.They both dissociate completely. Always write the major species that exist Always write the major species that exist

beforebefore any reaction occurs!! any reaction occurs!! Let’s titrate 50.0 mL of 0.200 Let’s titrate 50.0 mL of 0.200 MM HNO HNO33

with 0.100 with 0.100 MM NaOH with the following NaOH with the following volumes (watch volume changes) . . .volumes (watch volume changes) . . .

Analyze the pH after adding total NaOH Analyze the pH after adding total NaOH to 0.0, 10.0, 20.0, 100.0, 150.0 ml . . .to 0.0, 10.0, 20.0, 100.0, 150.0 ml . . .

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Strong acid with Strong Base pp Titration of 50.0 mL of 0.200 Titration of 50.0 mL of 0.200 MM HNO HNO33 with with

0.000.00 of 0.100 of 0.100 MM NaOH What’s pH? NaOH What’s pH? Major species?Major species? HH1+1+ NO NO33

1-1- H H22O . . . Stoich reaction? . . .O . . . Stoich reaction? . . . Nope, so Nope, so pHpH = -log [H = -log [H1+1+] = -log[.2] = ] = -log[.2] = .699.699

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1010.0 mL of 0.100 .0 mL of 0.100 MM NaOH What’s pH? NaOH What’s pH? Major species?Major species? HH1+1+ NO NO33

1-1- H H22O NaO Na1+1+ OHOH1-1- . . . Reaction?. . . Reaction?

RR HH1+1+ + + ++ OH OH1-1- H H22O , do “RIF”O , do “RIF” II 10 10 mmolmmol 1 1 mmolmmol (note: mmol, not molarity)(note: mmol, not molarity)

FF 9 9 0 0 [H[H1+1+] = 9 mmol/how much solution?] = 9 mmol/how much solution? [H[H1+1+] = 9 mmol/] = 9 mmol/6060 mL = 0.15 mL = 0.15 MM, pH = ??, pH = ?? pH = 0.82pH = 0.82

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Strong acid with Strong Base pp Titration of 50.0 mL of 0.200 M HNO3 with

20.0 of 0.100 M NaOH What’s pH? Major species = H1+ NO3

1- H2O Na1+ OH1-

R H1+ + OH1- H2O , do “RIF” I 10 mmol 2.0 mmol

F 8 0 [H1+] = 8 mmol/70 mL = 0.11 M pH = 0.942 Note: the volume is different from when

we previously added only 10.0 mL NaOH

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100100.0 of 0.100 .0 of 0.100 MM NaOH What’s pH? NaOH What’s pH? Major species = HMajor species = H1+1+ NO NO33

1-1- H H22O NaO Na1+1+ OH OH1- 1-

RR HH1+1+ + + ++ OH OH1-1- H H22O , do “RIF”O , do “RIF” II 10 10 mmolmmol 10 10 mmolmmol

FF 00 0 0 Major species now = NOMajor species now = NO33

1-1- H H22O NaO Na1+1+ Only =m is HOnly =m is H22OO H H1+1+ + + ++ OH OH1-1-

pH = 7.00pH = 7.00

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150150.0 of 0.100 .0 of 0.100 MM NaOH What’s pH? NaOH What’s pH? Major species = HMajor species = H1+1+ NO NO33

1-1- H H22O NaO Na1+1+ OH OH1- 1- RR HH1+1+ + + ++ OH OH1-1- H H22O , do “RIF”O , do “RIF”

II 10 10 mmolmmol 15 15 mmolmmol FF 00 5 5

Major species now = NOMajor species now = NO331-1- H H22O NaO Na1+1+ OHOH1-1-

[ OH[ OH1-1-] = 5 mmol/200ml = 0.025 ] = 5 mmol/200ml = 0.025 MM, pH = ?, pH = ? pH = 12.40pH = 12.40 (remember to subtract p (remember to subtract pOHOH

from 14.00!!) from 14.00!!)

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pH

mL of Base added

7

Strong acid with strong Base pp

Equivalence at pH 7

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Weak acid with Strong base pp

There There isis an equilibrium. an equilibrium. Do stoichiometry Do stoichiometry firstfirst.. ThenThen do equilibrium. do equilibrium. Titrate 50.0 mL of 0.10 Titrate 50.0 mL of 0.10 MM HAc HAc

(Ka = 1.8 x 10(Ka = 1.8 x 10-5-5) with 0.10 ) with 0.10 MM NaOH adding to a total addition volume NaOH adding to a total addition volume of 0.0, 10.0, 25.0, 50.0 and 60.0 ml of 0.0, 10.0, 25.0, 50.0 and 60.0 ml sequentially.sequentially.

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Weak acid with Strong base pp Titrate 50.0 mL of 0.10 Titrate 50.0 mL of 0.10 MM HAc HAc

with with 0.00 mL0.00 mL of 0.10 of 0.10 MM NaOH. NaOH. Major species?Major species? HAc HHAc H22O, Stoich???O, Stoich??? No. So go to “RICE”No. So go to “RICE” RR HAc HAc HH1+1+ + Ac + Ac1-1- I I

.10 .10 MM 00 0 0 CC - x- x +x+x +x +x EE.10 - x.10 - x xx xx

pH = ???pH = ??? pH = 2.87pH = 2.87

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with with 10.00 mL10.00 mL of 0.10 of 0.10 MM NaOH. NaOH. Major species?Major species? HAc NaHAc Na1+1+ OHOH1-1- H H22O, Stoich???O, Stoich??? Yes. Yes. Strong base & weak acidStrong base & weak acid. “RIF”. “RIF” RR HAc HAc + OH+ OH1-1- HOH + Ac HOH + Ac1-1-

I I 5 5 mmolmmol 1 1 mmolmmol 0 0 FF 44 0 0 1 1

What are major species now?What are major species now?

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Weak acid with Strong base pp HAc NaHAc Na1+1+ Ac Ac1-1- H H22O, Stoich???O, Stoich??? NoNo. No Strong base. RICE or H-H eq. No Strong base. RICE or H-H eq RR HAc HAc + + HH1+1+ + Ac + Ac1-1- I I

4/4/6060 00 1/ 1/6060 CC - x - x + x+ x + x + x E E(4/60) - x(4/60) - x + x+ x (1/60) + x (1/60) + x

Simplifying: (x)(1/60)/(4/60) = 1.8 x 10Simplifying: (x)(1/60)/(4/60) = 1.8 x 10-5-5 pH = ???pH = ??? pH = 4.14pH = 4.14 H/H: H/H: pH =pH = pKa + log([Ac pKa + log([Ac--]/[HAc]) = ]/[HAc]) = 4.144.14

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with with 25.0025.00 ml of 0.10 ml of 0.10 MM NaOH. NaOH. Major species?Major species? HAc NaHAc Na1+1+ OH OH1-1- H H22O, Stoich???O, Stoich??? Yes. Strong base & weak acid. “RIF”Yes. Strong base & weak acid. “RIF” RR HAc HAc + OH+ OH1-1- HOH + Ac HOH + Ac1-1-

I I 5 5 mmolmmol 2.5 2.5 mmolmmol 0 0 FF 2.52.5 0 0 2.5 2.5

What are major species now? . . .What are major species now? . . .

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Weak acid with Strong base pp HAc NaHAc Na1+1+ Ac Ac1-1- H H22O, Stoich???O, Stoich??? No.No. No Strong base. RICE or H-H eq No Strong base. RICE or H-H eq RR HAc HAc + + HH1+1+ + Ac + Ac1-1- I I

2.5/2.5/7575 00 2.5/ 2.5/7575 CC - x - x + x+ x +x +x

Can use RICE or H-H or . . .Can use RICE or H-H or . . . Since [HAc] = [AcSince [HAc] = [Ac1-1-], pH = pKA ], pH = pKA pH = ???pH = ??? pH = 4.74pH = 4.74

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with with 50.00 ml50.00 ml of 0.10 of 0.10 MM NaOH. NaOH. Major species?Major species? HAc NaHAc Na1+1+ OH OH1-1- H H22O, Stoich???O, Stoich??? YesYes. Strong base & weak acid. “RIF”. Strong base & weak acid. “RIF” RR HAc HAc + OH+ OH1-1- HOH + Ac HOH + Ac1-1-



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Weak acid with Strong base pp NaNa1+1+ Ac Ac1-1- H H22O, Stoich???O, Stoich??? NoNo. No Strong base. RICE or H-H eq. No Strong base. RICE or H-H eq What is the =m reaction & RICE?What is the =m reaction & RICE? RR Ac Ac1-1- + H + H22OO HAc + OHHAc + OH1-1-

I I 5/5/100100 0 0 0 0 CC - x - x + x+ x +x +x EE .05 - x .05 - x MM x x x x

xx22/(.05) = k/(.05) = kbb = 10 = 10-14-14/1.8 x 10/1.8 x 10-5-5 pH = ???pH = ??? pH = 8.72pH = 8.72 (14 - pOH) (14 - pOH)

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with with 60.00 mL60.00 mL of 0.10 of 0.10 MM NaOH. NaOH. Major species?Major species? HAc NaHAc Na1+1+ OH OH1-1- H H22O, Stoich???O, Stoich??? Yes. Strong base & weak acid. “RIF”Yes. Strong base & weak acid. “RIF” RR HAc HAc + OH+ OH1-1- HOH + Ac HOH + Ac1-1-



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Weak acid with Strong base pp

NaNa1+1+ AcAc1-1- H H22O, O, OHOH1-1- Have 2 bases, but OHHave 2 bases, but OH1-1- is stronger. is stronger. Amount OHAmount OH1-1- from Ac from Ac1-1- & H & H22O is small O is small

(negligible compared to that from NaOH).(negligible compared to that from NaOH). [OH[OH1-1-] = 1.0 mmol/110 ml] = 1.0 mmol/110 ml pH = ???pH = ??? pH = 11.96pH = 11.96 (14 - pOH) (14 - pOH)

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Diprotic Acid pp

Calculate [HCalculate [H++] after 75.0 ml of 0.20 ] after 75.0 ml of 0.20 M M NaOH has been added to 100.0 ml of NaOH has been added to 100.0 ml of 0.10 0.10 MM H H22AA

KaKa11 = 1.5 x 10 = 1.5 x 10-4-4

KaKa22 = 8.0 x 10 = 8.0 x 10-7-7 Steps . . .Steps . . . Treat each HTreat each H1+1+ as being separately as being separately

reacted with OHreacted with OH1-1-

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Diprotic Acid pp [H[H++] with 75.0 ml of 0.20 ] with 75.0 ml of 0.20 M M NaOH + 100.0 ml of NaOH + 100.0 ml of

0.10 0.10 MM H H22A (KaA (Ka11 = 1.5 x 10 = 1.5 x 10-4-4; Ka; Ka22 = 8.0 x 10 = 8.0 x 10-7-7). ).

RR HH22A + OHA + OH1- 1- HAHA1-1- + HOH + HOH

II 10 15 10 15

FF 0 5 0 5 1010 RR HAHA1-1- + OH + OH1-1- A A2-2- + HOH + HOH

II 10 5 010 5 0

FF 5 0 55 0 5 RR HAHA1-1- HH1+1+ + A + A2-2-

ICEICE 5-x 5-x x x 5+x . . .5+x . . .

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Diprotic Acid pp [H[H++] with 75.0 ml of 0.20 ] with 75.0 ml of 0.20 M M NaOH + 100.0 ml of NaOH + 100.0 ml of

0.10 0.10 MM H H22A (KaA (Ka11 = 1.5 x 10 = 1.5 x 10-4-4; Ka; Ka22 = 8.0 x 10 = 8.0 x 10-7-7).).

RR HAHA1-1- HH1+1+ + A + A2-2- ICEICE 5 5 x x 55

Since [HASince [HA1-1-] = [A] = [A2-2-], pH = pKa], pH = pKa22 = 6.10 = 6.10 You will have a problem like this on the test!You will have a problem like this on the test!

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Titration Curves

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pH

mL of Base added

7

Strong acid with strong Base Equivalence at pH 7

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Titration Curve: Strong Acid & Strong BaseTitration Curve: Strong Acid & Strong BaseOver what pH Over what pH

range does the range does the titrated solution titrated solution suddenly change suddenly change from high [Hfrom high [H33OO1+1+] ] to high [OHto high [OH1-1-]]

About 2.5 to 12About 2.5 to 12Look at page 756 Look at page 756

to find the best to find the best indicator . . .indicator . . .

Bromothymol Blue Bromothymol Blue (its midpoint is (its midpoint is close to the close to the equivalent point)equivalent point)

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pH

mL of Base added

>7

Weak acid with strong Base Equivalence at pH >7

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Titration Curve for Weak Acid & Strong BaseTitration Curve for Weak Acid & Strong BaseOver what pH range Over what pH range does the titrated does the titrated solution suddenly solution suddenly change from high change from high [H[H33OO1+1+] to high ] to high [OH[OH1-1-]?]?About 6.5 to 11About 6.5 to 11What’s pH when What’s pH when 0.067 mol NaOH 0.067 mol NaOH added?added?pH = 5pH = 5

What’s [HWhat’s [H33OO1+1+]?]?

[H[H33OO1+1+] = 1 x 10] = 1 x 10-5-5 MMBest Indicator (p. Best Indicator (p. 756 & next slide)?756 & next slide)?oo-Cresolphthalein or -Cresolphthalein or Phenolphthalein.Phenolphthalein.

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Useful pH ranges for several common indicators.Useful pH ranges for several common indicators.

p. 715

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Equivalence point (endpoint)• Point at which equal amounts Point at which equal amounts

of Hof H33OO++ and OH and OH-- have been have been

added.added.• Determined by…Determined by…

• indicator color changeindicator color change

TitrationTitration

• dramatic change in pHdramatic change in pH• pKpKaa is 1/2-way point is 1/2-way point

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pH

mL of Base added

7

Strong base with strong acid Equivalence at pH 7

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pH

mL of Base added

<7

Weak base with strong acid Equivalence at pH <7

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Summary - titrating Summary - titrating acidsacids pp

Strong acid and base just stoichiometry.Strong acid and base just stoichiometry. Weak acids: determine KWeak acids: determine Kaa, use for 0 mL base , use for 0 mL base

addedadded Weak acid Weak acid beforebefore equivalence point equivalence point

–Stoichiometry firstStoichiometry first–Then Henderson-Hasselbach or RICEThen Henderson-Hasselbach or RICE

No weak acid No weak acid atat equivalence point: use K equivalence point: use Kbb

and get pH by 14 - pOHand get pH by 14 - pOH Have weak base Have weak base afterafter equivalence, but use equivalence, but use

leftover leftover strongstrong base since it predominates. base since it predominates.

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Summary - titrating Summary - titrating basesbases pp

Determine KDetermine Kbb, use for 0 mL acid added., use for 0 mL acid added. Calc. pOH, then subtract from 14 for pHCalc. pOH, then subtract from 14 for pH Weak Weak basebase beforebefore equivalence point. equivalence point.

–Stoichiometry firstStoichiometry first–Then Henderson-HasselbachThen Henderson-Hasselbach–Subtract pOH from 14 for pHSubtract pOH from 14 for pH

No weak No weak basebase atat equivalence point; so use equivalence point; so use KKaa..

Have weak acid Have weak acid afterafter equivalence, but use equivalence, but use leftover leftover strongstrong acid since it predominates. acid since it predominates.

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15.5 Acid/Base Indicators Weak acids that change color when they Weak acids that change color when they

become bases.become bases. weak acid written HInweak acid written HIn HInHIn H H++ + In + In--

clearclear red red Equilibrium is controlled by pHEquilibrium is controlled by pH End point - when the indicator changes End point - when the indicator changes

color.color. Transition interval - pH range over which Transition interval - pH range over which

an indicator changes color. See next slide.an indicator changes color. See next slide.

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Figure 15.8 p. 756: The Useful pH Ranges for

Several Common Indicators

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Color Ranges of Various Indicators Used in TitrationsColor Ranges of Various Indicators Used in Titrations

Why are different indicators Why are different indicators used for SA-SB, SA-WB, or used for SA-SB, SA-WB, or WA-SB titrations?WA-SB titrations?

Each type of titration has a Each type of titration has a different end point.different end point.

Is it better to use an Is it better to use an indicator with a color indicator with a color change over a wide or change over a wide or narrow pH range. Why?narrow pH range. Why?

Narrow pH range since Narrow pH range since gives better approximation gives better approximation of the end point.of the end point.

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pH Scale RelationshipspH Scale Relationships

If [HIf [H33OO1+1+] of “x” is 10 x greater than another with pH 3, what is pH of “x”?] of “x” is 10 x greater than another with pH 3, what is pH of “x”?

pH = 2pH = 2If [OHIf [OH1-1-] of “y” is 1000 x greater than pure water’s, what is pH of “y”?] of “y” is 1000 x greater than pure water’s, what is pH of “y”?pH = 10pH = 10Is hand soap more or less basic than baking soda?Is hand soap more or less basic than baking soda?More basic.More basic.

63

Figure 15.9 p. 757Figure 15.9 p. 757

The pH Curve for the The pH Curve for the Titration of 100.0 mL of Titration of 100.0 mL of 0.10 0.10 MM HCI with 0.10 HCI with 0.10 MM NaOHNaOH

Which is the better Which is the better indicator?indicator?

Neither, since their Neither, since their ranges are not close to ranges are not close to the equivalence (end) the equivalence (end) point.point.

64

Figure 15.10 p. 757Figure 15.10 p. 757

The pH Curve for the The pH Curve for the Titration of 50 mL of Titration of 50 mL of 0.1 0.1 MM HC HC22HH33OO22 with with

0.1 0.1 MM NaOH NaOH

Choose the best Choose the best indicator between the indicator between the two . . .two . . .

PhenolphthaleinPhenolphthalein

65

Indicators pp

Moving towards equilibrium causes a Moving towards equilibrium causes a gradual gradual color changecolor change..

It is It is noticeablenoticeable when the when the ratioratio of of [In[In--]/[HI] or ]/[HI] or [HI]/[In[HI]/[In--] is 1:10 (] is 1:10 (1/101/10))

The Indicator is a The Indicator is a weak acidweak acid, so it has a , so it has a KKaa.. The The pHpH the indicator changes at is when the indicator changes at is when

[In [In--]/[HI] = ]/[HI] = 1/101/10. Using H-H . . .. Using H-H . . . That pHThat pH = pKa +log([In = pKa +log([In--]/[]/[HIHI]) = pKa +log(1/10) ]) = pKa +log(1/10) So, pH at which the indicator changes = pKa So, pH at which the indicator changes = pKa -- 1 1

on the way on the way upup the pH scale (i.e., titrating an acid the pH scale (i.e., titrating an acid using a base).using a base).

66

Indicators pp

pH=pKa + log([pH=pKa + log([HIHI]/[In]/[In--]) = pKa + log(10)]) = pKa + log(10) pH = pKpH = pKaa ++ 1 on the way 1 on the way downdown (i.e., (i.e.,

titrating a base by adding an acid).titrating a base by adding an acid). Choose the indicator with a pKa 1 Choose the indicator with a pKa 1 lessless

than the pH at equivalence point if you than the pH at equivalence point if you are are titrating an acid with basetitrating an acid with base..

Choose the indicator with a pKa 1 Choose the indicator with a pKa 1 greatergreater than the pH at equivalence point than the pH at equivalence point if you are if you are titrating a base with acidtitrating a base with acid..

67

15.6 Solubility Equilibria Will it all dissolve, and if not, how much?Will it all dissolve, and if not, how much? Note on this week’s online HW:Note on this week’s online HW: One or more questions contains more One or more questions contains more

information than you need.information than you need. You have to discriminate between what You have to discriminate between what

information is relevant.information is relevant. Strategy: Given all the information figure Strategy: Given all the information figure

out your plan and discard information not out your plan and discard information not needed for that plan.needed for that plan.

Have fun!Have fun!

68

AllAll dissolving is an dissolving is an equilibriumequilibrium.. If there is not much solid it will all dissolve.If there is not much solid it will all dissolve. As more solid is added the solution will As more solid is added the solution will

become saturated.become saturated. SolidSolid dissolveddissolved Solid will precipitate as fast as it dissolves.Solid will precipitate as fast as it dissolves. So, it is a So, it is a dynamic equilibriumdynamic equilibrium RememberRemember: pure liquids and pure solids : pure liquids and pure solids

are never included in an equilibrium are never included in an equilibrium expression.expression.

Solubility Equilibria

69

General equation MM++ stands for the cation (usually metal). stands for the cation (usually metal). NmNm-- stands for the anion (a nonmetal).stands for the anion (a nonmetal). MMaaNmNmbb(s) (s) aaMM++(aq) + (aq) + bbNmNm- - (aq) (aq) K = [MK = [M++]]aa[Nm[Nm--]]bb/[M/[MaaNmNmbb]] But the concentration of a solid doesn’t change But the concentration of a solid doesn’t change

(so drops out of the =m expression).(so drops out of the =m expression). KKspsp = [M = [M++]]aa[Nm[Nm--]]bb

This is called the This is called the solubility productsolubility product for each for each compound.compound.

70

Watch out SolubilitySolubility is is notnot the same as the same as solubility solubility

productproduct.. Solubility productSolubility product is an is an equilibrium equilibrium

constantconstant.. It does NOT change (It does NOT change (exceptexcept with with

temperature).temperature). SolubSolubilityility is an equilibrium is an equilibrium positionposition for for

how much can dissolve (so can shift).how much can dissolve (so can shift). A A common ioncommon ion can change can change solubilitysolubility..

71

Calculating Calculating KKspsp The The solubilitysolubility of iron(II) oxalate FeC of iron(II) oxalate FeC22OO4 4

is 65.9 is 65.9 mgmg/L /L The solubility of LiThe solubility of Li22COCO33 is 5.48 is 5.48 gg/L/L

We would use the solubility to then We would use the solubility to then calculate calculate KKspsp for each of them. for each of them.

72

Calculating Solubility pp

The solubility is determined by the The solubility is determined by the equilibrium.equilibrium.

It is an equilibrium problem (RICE).It is an equilibrium problem (RICE). Calculate the Calculate the KKspsp of Bi of Bi22SS33, with a , with a solubilitysolubility

of of 1.0 x 101.0 x 10-15 -15 mol/Lmol/L @ 25 @ 25ooC. Steps . . .C. Steps . . . Major species???Major species??? BiBi22SS33, H, H22OO RICE is what? . . .RICE is what? . . .

73

Calculating Ksp pp

KKspsp of Bi of Bi22SS33, with , with solubilitysolubility of 1.0 x 10 of 1.0 x 10-15 -15

mol/L @ 25mol/L @ 25ooC.C. Solubility means the Solubility means the amount that amount that

dissolvesdissolves.. RR Bi Bi22SS33 22BiBi3+3+ + + 33SS2-2- II 11.0 x 10.0 x 10-15-15 MM 00 0 0

CC - -1.0 x 101.0 x 10-15-15 + +22(1.0 x 10(1.0 x 10-15-15) ) 33(1.0 x 10(1.0 x 10-15-15)) E E 00 22.0 x 10.0 x 10-15-15 3 3.0 x 10.0 x 10-15-15

KKspsp = [Bi = [Bi3+3+]]22[S[S2-2-]]33 = ( = (22.0 x 10.0 x 10-15-15))22 ((33.0 x 10.0 x 10-15-15))33

KKspsp = 1.1 x 10 = 1.1 x 10-73-73 (very small number) (very small number)

74

Calculating Solubility Calculating Solubility from Kfrom Kspsp pp

Calculate the solubility in Calculate the solubility in gg/L of Cu(IO/L of Cu(IO33))22

with a Kwith a Kspsp of 1.4 x 10 of 1.4 x 10-7 -7 MM. Steps . . .. Steps . . . Major species . . .Major species . . . Cu(IOCu(IO33))22 & H & H22O. Now do RICEO. Now do RICE

RR Cu(IO Cu(IO33))22 Cu Cu2+2+ + 2 IO + 2 IO331-1-

II xx 00 0 0 CC -x -x +x +x + +22x x E E 00 xx 2x 2x

75

Calculating Solubility from Ksp pp

Calculate the solubility in Calculate the solubility in gg/L of Cu(IO/L of Cu(IO33))22 (M (M

= 413g/mol)= 413g/mol) with a K with a Kspsp of 1.4 x 10 of 1.4 x 10-7-7 MM. . RR Cu(IO Cu(IO33))22 Cu Cu2+2+ + 2 IO + 2 IO33

1-1-

II xx 00 0 0

CC -x -x +x +x + +22x x E E 00 xx 2x 2x

KKspsp = 1.4 x 10 = 1.4 x 10-7-7 = [x][2x] = [x][2x]22 = 4x = 4x33 = ??? = ??? x = 3.3 x 10x = 3.3 x 10-3-3 molmol/L = 1.4 /L = 1.4 gg/L/L

76

Relative solubilities pp

KKspsp alonealone will only allow us to compare the will only allow us to compare the

solubility of solubility of differentdifferent solids that fall apart solids that fall apart into the into the samesame number of ions. number of ions.

E.g.,E.g., if asked to compare the if asked to compare the AgIAgI, , CuICuI and and CaSOCaSO44 solubilitiessolubilities, compare K, compare Kspsp values values

((1.5 x 101.5 x 10-16-16, , 5.0 x 105.0 x 10-12-12, , 6.1 x 106.1 x 10-5-5)) Biggest KBiggest Kspsp is most soluble ( is most soluble (CaSOCaSO44). ). We can do this We can do this ONLYONLY because they fall because they fall

apart into the apart into the samesame number of ions. number of ions.

77

Relative solubilities pp

If they fall apart into If they fall apart into differentdifferent number of pieces number of pieces you have to do the math.you have to do the math.

E.gE.g., CuS, Ag., CuS, Ag22S & BiS & Bi22SS33 SeeSee Table 15.5 p. 722. Table 15.5 p. 722.

SaltSalt KKspsp Solubility @ 25Solubility @ 25oo C C CuS CuS

8.5 x 108.5 x 10-45-45 9.2 x 109.2 x 10-23-23 mol/L Ag mol/L Ag22SS 1.6 x 1.6 x

1010-49-49 3.4 x 103.4 x 10-17-17 mol/L Bi mol/L Bi22SS33 1.1 x 101.1 x 10--7575

1.0 x 101.0 x 10--1515 mol/L mol/L Even though BiEven though Bi22SS33 has the least K has the least Kspsp, it has , it has

greatest solubility!greatest solubility! Watch for trick AP questions on this.Watch for trick AP questions on this.

78

Common Ion Effect pp

If we try to dissolve the solid in a If we try to dissolve the solid in a solution with solution with eithereither the cation or anion the cation or anion already presentalready present lessless will dissolve. will dissolve.

This is Le Chatelier’s principle.This is Le Chatelier’s principle. This is why you This is why you mustmust consider initial consider initial

concentrations (major species)!concentrations (major species)! Calculate the solubility of Calculate the solubility of solidsolid Ca CaFF22 (K (Kspsp

= 4.0 x 10= 4.0 x 10-11-11) ) inin a 0.025 a 0.025 MM Na NaFF solution. solution. Steps. . .Steps. . .

79


CaCaFF2(s)2(s) solubility in 0.025 solubility in 0.025 MM Na NaFF solution. solution.

(K(Kspsp = 4.0 x 10 = 4.0 x 10-11-11) Major species???) Major species???

CaFCaF22 Na Na1+1+ F F1-1- H H22O. O. Write RICE (+ simplifying assumptions if Write RICE (+ simplifying assumptions if

KKspsp is <<) . . . is <<) . . .

RR Ca CaFF2(s)2(s) Ca Ca2+2+ + 2 F + 2 F1-1- II xx 00 0.025 0.025

CC -x -x +x +x 22x x E E 00 xx .025 + .025 +22x x

80


CaCaFF2(s)2(s) solubility in 0.025 solubility in 0.025 MM Na NaFF solution. solution. (K(Kspsp = 4.0 x 10 = 4.0 x 10-11-11)) RR Ca CaFF2(s)2(s) Ca Ca2+2+ + 2 F + 2 F1-1- II xx 00 0.025 0.025 CC -x -x +x +x 22x x E E 00 xx .025 + .025 +22x x

4.0 x 104.0 x 10-11-11 = (x)(.025 + 2x) = (x)(.025 + 2x)22 a cubic!!! a cubic!!! But, KBut, Kspsp = <<, so .025 + 2x ≈ 0.025 = <<, so .025 + 2x ≈ 0.025 4.0 x 104.0 x 10-11-11 = (x)(.025) = (x)(.025)22 x = 6.4 x 10x = 6.4 x 10-8-8 MM

81

pH and solubility OHOH-- can be a common ion. can be a common ion. E.g.,E.g., Mg(OH) Mg(OH)22 Mg Mg2+2+ + 2OH + 2OH1-1-

Adding OHAdding OH- - shifts shifts leftleft, so more soluble in , so more soluble in acid. This is how antacids work.acid. This is how antacids work.

Don’t want Mg(OH)Don’t want Mg(OH)22 too soluble or it too soluble or it tears up the stomach.tears up the stomach.

So, it only gets more soluble So, it only gets more soluble whenwhen there there is acid, which it then neutralizes (shifting is acid, which it then neutralizes (shifting right to do so) and the rest of it stays right to do so) and the rest of it stays insolubleinsoluble until needed or excreted. until needed or excreted.

82

pH and Solubility pp

Other Other anionsanions: If they come from a : If they come from a weak acid (weak acid (i.e.,i.e., are effective bases) are effective bases) their their saltsalt is is moremore soluble in soluble in acidicacidic solution than in water (shifting =m).solution than in water (shifting =m).

CaCCaC22OO44 Ca Ca+2+2 + C + C22OO44-2-2

HH++ + C + C22OO44-2 -2 HCHC22OO44

- -

HH++ reduces C reduces C22OO44-2-2 (acidic solution) (acidic solution)

so salt dissolution shifts so salt dissolution shifts rightright..

83

15.7 Precipitation & Qualitative Analysis pp

Ion Product, Q =[MIon Product, Q =[M++]]ooaa[Nm[Nm--]]oo

bb

If Q > KIf Q > Kspsp a precipitate forms. a precipitate forms.

If Q < KIf Q < Kspsp No precipitate. No precipitate.

If Q = KIf Q = Kspsp at equilibrium. at equilibrium.

A solution of 750.0 mL of 4.00 x 10A solution of 750.0 mL of 4.00 x 10-3-3MM Ce(NOCe(NO33))33 is added to 300.0 mL of is added to 300.0 mL of

2.00 x 102.00 x 10-2-2MM KIO KIO33. . Will Ce(IOWill Ce(IO33))33 (Ksp= (Ksp=

1.9 x 101.9 x 10-10-10MM)) precipitate and if so, what precipitate and if so, what is the concentration of the ions?is the concentration of the ions? Steps. Steps.

84

Precipitation & Qualitative Analysis pp

750.0 mL of 4.00 x 10750.0 mL of 4.00 x 10-3-3MM Ce(NO Ce(NO33))33

added to 300.0 mL of 2.00 x 10added to 300.0 mL of 2.00 x 10-2-2MM KIOKIO33. . Ksp= 1.9 x 10Ksp= 1.9 x 10-10 -10 MM. . Will Ce(IO Will Ce(IO33))33

ppt; if so, what is [ ] of the ions?ppt; if so, what is [ ] of the ions? Calculate [ions] in the Calculate [ions] in the mixedmixed solution solution

beforebefore any reaction. any reaction. Then calculate Q & compare to KThen calculate Q & compare to Kspsp. .

Steps follow . . .Steps follow . . .

85


750.0 mL of 4.00 x 10-3M Ce(NO3)3 + 300.0 mL of 2.00 x 10-2M KIO3. Ksp=

1.9 x 10-10 M. (M = mol/L)

[Ce3+]o = (750 mL)(4 x 10-3 mmol/ml) (750 mL + 300 mL)

[Ce3+]o = 2.86 x 10-3 M

[IO31-]o = 5.71 x 10-3 M (same method)

86


Will Ce(IOWill Ce(IO33))33 ppt? ppt? KKspsp = 1.9 x 10 = 1.9 x 10-10-10

Ce(IOCe(IO33))33 CeCe3+3+ + 3IO + 3IO331-1-

[Ce[Ce3+3+]]oo = 2.86 x 10 = 2.86 x 10--3 3 MM [IO[IO33

1-1-]]oo = 5.71 x 10 = 5.71 x 10-3-3 MM (same method) (same method) Q = (2.86 x 10Q = (2.86 x 10-3-3)(5.71 x 10)(5.71 x 10-3-3))33 Why isn’t [IOWhy isn’t [IO33

1-1-] ] multipliedmultiplied by 3 also? by 3 also? We already have an actual [ ] for it.We already have an actual [ ] for it. Q = 5.32 x 10Q = 5.32 x 10-10-10 >> K Kspsp = 1.9 x 10 = 1.9 x 10-10-10

Ppt forms! Ppt forms! What are the =m ion [ ]s?What are the =m ion [ ]s?

87


What are =m ion [ ]s?[Ce3+]o = 2.86 x 10-3 M; [IO3

1-]o = 5.71 x 10-3 M Do stoich, then backtrack to =m R Ce3+ + 3IO3

1- Ce(IO3)3

I (2.86 x 10-3 M)(1050 mL) (5.71 x 10-3 M)(1050 mL) = 3.00 mmol 6.00 mmol F 1.00 mmol 0 mmol (remember mole ratios & limiting

reactant) . . .

88


What are What are =m=m ion [ ]s cont. ion [ ]s cont. RR Ce Ce3+3+ + + 3IO3IO33

1-1- Ce(IOCe(IO33))33

FF 1 mmol 1 mmol 0 mmol0 mmol (2 mmol)(2 mmol) Now do =m (RICE) - Now do =m (RICE) - “Backtrack” the reaction“Backtrack” the reaction RR Ce(IOCe(IO33))33 CeCe3+3+ + + 3IO3IO33

1-1-

II x x 1 mmol/1050 mL 1 mmol/1050 mL 0 0 CC -x -x +x+x + 3x+ 3x EE (1 mmol + x) /(1050 mL (1 mmol + x) /(1050 mL

0 0 ≈ 9.52 x 10 ≈ 9.52 x 10-4-4 3x 3x We know that We know that KKspsp = 1.9 x 10 = 1.9 x 10-10-10

Solve for 3x = [IOSolve for 3x = [IO331-1-] = (next slide)] = (next slide)

89


What are What are =m ion=m ion [ ]s cont. [ ]s cont. RR Ce(IOCe(IO33))33 CeCe3+3+ + + 33IOIO33

1-1-

I I x x 1 mmol/1050 mL 1 mmol/1050 mL 0 0 CC -x -x +x+x + 3x+ 3x EE (1 mmol/(1050 mL) + x (1 mmol/(1050 mL) + x

9.52x109.52x10-4-4 + + xx 33xxKKspsp = 1.9x10 = 1.9x10-10-10 solve for x, then get 3x solve for x, then get 3x

1.9 x 101.9 x 10-10-10 = (9.52 x 10 = (9.52 x 10-4-4)(3x))(3x)33 = (9.52 x 10 = (9.52 x 10-4-4)()(2727xx33)) xx = = 1.95 x 101.95 x 10-3-3

3x = [3x = [IOIO331-1-] = 5.84 x 10] = 5.84 x 10-3-3

[Ce[Ce3+3+] = ] = 2.90 x 102.90 x 10-3-3 (= (= 9.52 x 109.52 x 10-4-4 + + 1.95 x 101.95 x 10-3-3) )

90

Selective Precipitation pp

• A solution has 1.0 x 10A solution has 1.0 x 10-4-4 MM CuCu11++ & 2.0 x 10 & 2.0 x 10-3-3 MM PbPb2+2+ If IIf I1-1- is added which is added which ppts 1st . . . ppts 1st . . . PbIPbI22 (K (Kspsp 1.4 x 10 1.4 x 10-8-8))

or or CuI (KCuI (Kspsp = 5.3 x 10 = 5.3 x 10-12-12))?? How much How much II1-1- is needed to is needed to

ppt each salt?ppt each salt? Steps. Steps.

• Need Q < KNeed Q < Kspsp for no ppt. So, when Q = K for no ppt. So, when Q = Kspsp that is that is

the maximum amount that can solubilize.the maximum amount that can solubilize.

• We know [CuWe know [Cu1+1+] & [Pb] & [Pb2+2+] & the above K] & the above Kspsp for both for both

insoluble compounds, so can calculate [Iinsoluble compounds, so can calculate [I1-1-] that ] that would put [Cuwould put [Cu1+1+] & [Pb] & [Pb2+2+] into a ppt.] into a ppt.

• 1.4 x 101.4 x 10-8-8 = [Pb = [Pb2+2+][I][I1-1-]]22 = (2.0 x 10 = (2.0 x 10-3-3)[I)[I1-1-]]22 [I[I1-1-] = 2.6 x 10] = 2.6 x 10-3-3 MM with Pb with Pb2+2+

• 5.3 x 105.3 x 10-12-12 = [Cu = [Cu1+1+][I][I1-1-] = (1.0 x 10] = (1.0 x 10-4-4)[I)[I1-1-] ] [I[I1-1-] = 5.3 x 10] = 5.3 x 10-8-8 MM with Cu with Cu1+1+

• So, CuI will precipitate out first.So, CuI will precipitate out first.

91

Selective Precipitations Used to separate mixtures of metal ions Used to separate mixtures of metal ions

in solutions.in solutions. Add anions that will only precipitate Add anions that will only precipitate

certain metals at a time.certain metals at a time.

Often use HOften use H22S because in S because in acidicacidic solution solution

the less soluble sulfides like Hgthe less soluble sulfides like Hg+2+2, Cd, Cd+2+2, ,

BiBi+3+3, Cu, Cu+2+2, Sn, Sn+4+4 will precipitate first, and will precipitate first, and

in basic solutions . . .in basic solutions . . .

92

Selective Precipitation In Basic (adding OHIn Basic (adding OH--)) solution, Ssolution, S-2-2 (from (from

HH22S) will increase so the S) will increase so the more solublemore soluble

sulfides will precipitate. (E.g., Cosulfides will precipitate. (E.g., Co+2+2, , ZnZn+2+2, Mn, Mn+2+2, Ni, Ni+2+2, Fe, Fe+2+2, Cr(OH), Cr(OH)33, ,

Al(OH)Al(OH)3)3)

I.e.,I.e., can regulate by changing the pH can regulate by changing the pH

93

Z5e p. 771 Fig 15.11 Precipitation by H2S

94

Selective precipitation Follow the steps first with insoluble Follow the steps first with insoluble

chlorideschlorides (Ag, Pb, Ba) (Ag, Pb, Ba) Then Then sulfidessulfides in in acidacid.. Then Then sulfidessulfides in in basebase.. Then insoluble Then insoluble carbonatecarbonate (Ca, Ba, Mg) (Ca, Ba, Mg) Alkali metals and NHAlkali metals and NH44

++ remain in remain in solution (solubility song).solution (solubility song).

See, flow chart figures 15.12 p. 730See, flow chart figures 15.12 p. 730 as I show you how to read them.as I show you how to read them.

95

Z5e Fig 15.12 p771 Selective Precipitation of Common Ions

96

Z5e p. 779 Fig. 15.13 Separation of Gr 1 Ions

97

15.8 Complex Ion Equilibria pp

A A positively charged ionpositively charged ion surrounded by surrounded by ligandsligands..

Ligands: Ligands: Lewis basesLewis bases using their using their lone pairlone pair to to stabilizestabilize the charged metal ions. the charged metal ions.

CommonCommon: : NHNH33, H, H22O, ClO, Cl--,CN,CN--, SCN, SCN--, OH, OH1-1- Memorize these! Need for net reactions. Memorize these! Need for net reactions. Coordination numberCoordination number is the number of is the number of

attached ligands (attached ligands (usually doubleusually double the ion the ion chargecharge).).

Cu(NHCu(NH33))44+2+2 has a coordination # of has a coordination # of 44

98

The addition of each ligand has its own equilibrium

Usually the ligand is Usually the ligand is in large excessin large excess compared to the metal ion.compared to the metal ion.

And the individual K’s will be large, so And the individual K’s will be large, so we can treat them we can treat them as ifas if they go to they go to completioncompletion..

The complex ion will be the biggest ion The complex ion will be the biggest ion in solution.in solution.

99

Metal Ion Surrounded by Water Molecules

100

Complex Ion Problem pp

Calc. [Calc. [AgAg1+1+], [], [Ag(SAg(S22OO33))1-1-], & [], & [Ag(SAg(S22OO33))223-3-]] in in

mixturemixture of of 150.0 mL of 1.00 x 10150.0 mL of 1.00 x 10-3-3 MM AgNO AgNO33 & &

200.0 mL of 5.00 200.0 mL of 5.00 MM Na Na22SS22OO33

AgAg1+1+ + S + S22OO332-2- Ag(S Ag(S22OO33))1-1- KK1 1 = 7.4 x 10= 7.4 x 1088

Ag(SAg(S22OO33))1-1- + S + S22OO332-2- Ag(S Ag(S22OO33))22

33-- KK2 2 = 3.9 x 10= 3.9 x 1044

[ ] [ ] beforebefore rxn : [Ag rxn : [Ag1+1+]]oo = (150)(10 = (150)(10-3-3)/(150+200) = 4.29x10)/(150+200) = 4.29x10-4-4

Ditto for [SDitto for [S22OO332-2-]]oo . . . = . . . 2.86 . . . = . . . 2.86 MM

Since [SSince [S22OO332-2-]]oo >> [Ag >> [Ag1+1+]]oo and K and K11 & K & K22 are are largelarge, assume , assume

that both formation reactions go to completion.that both formation reactions go to completion. See “RIF” next slide.See “RIF” next slide. AgAg1+1+ is is limitinglimiting & << compared to S & << compared to S22OO33

2-2-

101

Complex Ion Problem cont. pp

R Ag1+ + 2S2O32- Ag(S2O3)2

3- I 4.29 x 10-4 M 2.86 M 0

F almost 0 2.86 - 2(4.29 x 10-4) 4.29 x 10-4 ≈ 2.86

Since there is some Ag1+ at =m & there is Ag(S2O3)1- in solution we calculate their [ ]s using K1 & K2

K2 = 3.9x104 = [Ag(S2O3)23-] [Ag(S2O3

)1-] = 3.8 x 10-9 M [Ag(S2O3)1-][2S2O3

2-]

K1 = 7.4x108 = [Ag(S2O3) 1-] [Ag1+] = 1.8 x 10-18 M

[Ag1+][S2O32-]

[Ag(S2O3)23-] >> [Ag(S2O3)1-] >> [Ag1+]

102

Strategies to dissolve a water-insoluble solid pp

If the If the anionanion is a good is a good basebase then increase then increase solubility by adding solubility by adding acidacid

E.g.,E.g., Mg(OH) Mg(OH)22 Mg Mg2+2+ + 2OH + 2OH--

If the anion is If the anion is notnot a good base, “ a good base, “complex itcomplex it” ” by dissolving in a solution that has a ligand by dissolving in a solution that has a ligand that forms complex ions with its that forms complex ions with its cationcation..

AgClAgCl((ss)) + 2NH + 2NH3(aq) 3(aq) Ag(NHAg(NH33))221+1+

((aqaq)) + Cl + Cl1-1-

(ac)(ac)

Aqua RegiaAqua Regia (HCl + HNO (HCl + HNO33) “royal water” ) “royal water” Dissolves lots of stuff, including many Dissolves lots of stuff, including many

sulfides. sulfides.

ch15 z5e aq. equil

Education

aweak acid

ac aqua regiahcl

salt oraweak base

solution thatresistsa

mnaac pp

mnaac ka

itreversesthe dissociation

complex ions