Dr. Steward’s CHM152 Exam Review 2 (17, 18, 19, 20) If an equal number of moles of HCN and KOH are added to water, what is the resulting solution: acidic, basic, or neutral? Explain. KOH is a strong base;; HCN is a weak acid. When reacted together, they’ll produce water and KCN. K+ and CN- which will react with water (it is a weak base, a conjugate of a weak acid). This will produce a basic solution Determine the pH of a buffer solution that contains 4.0M CH3COOH and 3.0M NaCH3COO. Ka = 1.8x10-5.

= 4.62

Determine the pH of the buffer solution from above after you add 7.00mL of 1.0M HCl to 50.00ml of the buffer solution. HCl + NaCH3COO → CH3COOH + NaCl 0.0070mol 0.15mol 0.20mol

-0.0070mol -0.0070mol +0.0070mol

0mol 0.143mol 0.207mol

= 4.58 Determine the pH of the original buffer solution after you add 4.00mL of 1.5M NaOH to 50.00ml of the buffer solution. NaOH + CH3COOH → NaCH3COO + H2O 0.0060mol 0.20mol 0.15mol

-0.0060mol -0.0060mol +0.0060mol = 4.64

0mol 0.194 0.156

Calculate the pH of a 0.100 M NaCH3COO solution. Ka for acetic acid is 1.8 x 10-5. (Use A- as CH3COO-):6 i A- + H2O HA + OH- 0.100 - 0 0 Kb = x2 / (0.100 – x) = 5.6 x 10-10

- x - + x + x Assume x << 0.100 (and check it!)

acid] [conjugatebase] [conjugate log pK pH a

4.0M3.0M log 4.74 pH

HCl mol0070.0HCl L

HCl 1.0molHCl L00700.0

COOHCH mol15.0COONaCH L

COONaCH 3.0molCOONaCH 0.05000L 3

3

33

COOHCH mol20.0COOHCH L

COOHCH 4.0molCOOHCH 0.05000L 3

3

33

0.207mol0.143mol log 4.74 pH

NaOH mol0060.0NaOH L

NaOH 1.5molNaOH L00400.0

0.194mol0.156mol log 4.74 pH

0.100 – x - x x Kb = x2 / 0.100 = 5.6 x 10-10 x2 = 5.6 x 10-11; x = 7.48 x 10-6; pOH = 5.13, pH = 8.87 Calculate the pH of a 0.100 M CH3NH3Cl solution. Kb for methylamine, CH3NH2, is 3.7 x 10-4. (Use BH+ as CH3NH3

+): BH+ + H2O H3O+ + B 0.100 - 0 0 Ka = x2 / (0.100 – x) = 2.7 x 10-11

- x - + x + x Assume x << 0.100 (and check it!) 0.100 – x - x x Ka = x2 / 0.100 = 2.7 x 10-11 x2 = 2.7 x 10-12; x = 1.64 x 10-6; pH = 5.78 A 50.0 mL sample of 0.500 M HC2H3O2 acid is titrated with 0.150 M NaOH. Ka = 1.8x10-5 for HC2H3O2. Calculate the pH of the solution after the following volumes of NaOH have been added: a) 0 mL; b) 166.7 mL; c) 180.0 mL. a) 0 ml of base; only a weak acid is initially present so [H+] ≠ [HA] HC2H3O2 H+ + C2H3O2

- I 0.500 0 0 C -x x x E 0.50-x x x

Ka = ][]][[

232

232

OHHCOHCH

1.8x10-5 = 5000

2

.x

[H +] = x = )x.(. 510815000 = 3.0x10-3 pH = -log 3.0x10-3 = 2.52

b) 166.7 ml of NaOH are added

moles HC2H3O2 =

L

OHHCmoles.mLL

ml. 232500010001

050 2.50x10-2 moles HC2H3O2

moles NaOH =

L

NaOHmoles.mLLml. 1500

100017166 2.50x10-2 moles NaOH

neutralization: HC2H3O2 + OH- C2H3O2- + H2O

I 0.0250 0.0250 0 C -0.0250 -0.0250 +0.0250 Final 0 0 0.0250

only acetate remains – a weak base:

[C2H3O2-] =

L.moles.

2167010502 2

0.115 M

base hydrolysis: C2H3O2- + H2O HC2H3O2 + OH-

I 0.115 0 0 C -x x x E 0.115-x x x

Kb for C2H3O2- = 5

14

108.1101

x

= 5.6x10-10 Kb = ][

]][[

232

232

OHCOHOHHC

5.6x10-10 =

115.0

2x

x = [OH-] = 10106.5115.0 = 8.0x10-6

pOH = -log 8.0x10-6 = 5.10 pH = 14 – 5.10 = 8.90 At the equivalence point for a WA/SB titration, the pH > 7 due to the OH-

produced by the conjugate base hydrolysis reaction.

c) 180.0 mL of NaOH are added from part b, moles HC2H3O2 = 2.50x10-2 moles HC2H3O2

moles NaOH =

L

NaOHmolesmLLml 150.0

1000100.180 2.70x10-2 moles NaOH

moles excess base = 2.70x10-2 moles - 2.50x10-2 moles = 2.0x10-3 moles NaOH

M OH- = M NaOH =

L.molesx.

230001002 3

8.7x10-3 M OH-

pOH = -log 8.7x10-3 = 2.06 pH = 14 – 2.06 = 11.94 *Excess NaOH remains - this is the primary source of OH-. We can neglect the hydrolysis of the conjugate base because this would contribute a relatively small amount of OH- compared to the amount that comes directly from the excess NaOH.

What is the concentration of a saturated silver acetate solution? Ksp(AgC2H3O2) = 1.94 x 10-3.

Since Ksp = [Ag+][C2H3O2-], and the concentration of silver ions is the same as the

concentration of acetate ions, we can set up the following equation: 1.94 x 10-3 = x2 x = 0.0440 M

What is the concentration of a saturated lead chloride solution? Ksp(PbCl2) = 1.17 x 10-5.

Ksp = [Pb+2][Cl-]2. Since the concentration of chloride ions is twice that of lead (II) ions, this boils down to the following equation:

1.17 x 10-5 = (x)(2x)2 1.17 x 10-5 = 4x3 Silver phosphate, Ag3PO4, is an insoluble salt that has a Ksp = 1.3 x 10-20.

a) Calculate the molar solubility of Ag3PO4 in pure water.

Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = [Ag+]3[PO4

3-] I 0 0 C 3x x E 3x x

Ksp = (3x)3x

1.3x10-20 = 27x4 x4 = 4.8x10-22 x = 4.7x10-6 M = molar solubility of Ag3PO4 in pure water

b) Calculate the molar solubility of Ag3PO4 in a solution containing 0.020 M Na3PO4 (a

soluble salt). soluble salt: Na3PO4 3Na+ + PO4

3- Phosphate is the common ion: [PO4

3-] = [Na3PO4] = 0.020 M (since 1 mol Na3PO4 forms 1 mol PO43- ions)

Ag3PO4(s) 3Ag+(aq) + PO43-(aq)

I 0 0.020 C 3x x E 3x 0.020+x Ksp = [Ag+]3[PO4

3-] 1.3x10-20 = = (3x)30.020

6.5x10-19 = 27x3 x3 = 2.4x10-20 x = 2.9x10-7M = molar solubility of Ag3PO4 with a common ion Adding common ion decreases the solubility of Ag3PO4 c) Will a precipitate form if a solution is made to contain 3.0x10-5M AgNO3 and 2.3x10-

4M Na3PO4? (Hint: Q vs. K)

We already know Ksp and Qsp = [Ag+]3[PO43-] so we can plug the concentrations in to

calculate Qsp. Qsp = [3.0x10-5]3[2.3x10-4] = 6.21x10-18. From above, Ksp = 1.3x10-20. Since Q>K, the reaction should shift to the reactant side (remember, the reaction was written with the solid as a reactant), and a solid will form. If Q had been smaller than K, a precipitate would not have formed.

a. Calculate the standard free energy change, ΔG, for the following at 25 C: MgO(s) + C(graphite) Mg(s) + CO(g) ΔH = 491.18 kJ ΔS = 197.67 J/K

G = H - TS

G = 491.18 kJ - (298 K)

KJ67.197

J

kJ10001 = + 432.27 kJ

b. Is this reaction spontaneous at 25 C? If not, at what temperature can we make this reaction spontaneous?

No, it is not spontaneous at 25 C (G is a large positive value)

Set G = 0: 0 = H - TS T = SH

T =

kJkJ

KJ

kJ

1000167.197

18.491 = 2484.8 K This is T at equilibrium, spont at T >

2484.8 K

For the unbalanced reaction 2 SO2(g) + O2(g) → 2 SO3(g) calculate ΔG at 25.0ºC when the reactants and product are at the following partial pressures: 10.0 atm SO2 , 10.0 atm O2 , and 1.00 atm SO3. G = Go + RT ln Q Q = 12 / (102 * 10) = 1/1000 Go = [ 2mol(-371.1 kJ/mol)] – [ 2mol(-300.2 kJ/mol) + 0 ] = -141.8 kJ G = -141.8 kJ + (8.314 J/mol·K) (298 K) ( kJ/1000J) (ln 1/1000) = -159 kJ

In each of the following equations, indicate the element that has been oxidized and the one that has been reduced. You should also label the oxidation state of each before and after the process: 2 Na + FeCl2 2 NaCl + Fe

Sodium is oxidized, going from a 0 to +1 oxidation state. Iron is reduced, going from a +2 to 0 oxidation state.

2 C2H2 + 5 O2 4 CO2 + 2 H2O

Carbon is oxidized, going from a –1 to +4 oxidation state. Oxygen is reduced, going from a 0 to –2 oxidation state.

2 PbS + 3 O2 2 SO2 + 2 PbO

Sulfur is oxidized, going from a –2 to +4 oxidation state. Oxygen is reduced, going from a 0 to –2 oxidation state.

A galvanic cell is constructed based on the following reactions:

Zn2+ + 2e- Zn E red = -0.76 V

Al3+ + 3e- Al E red = -1.66 V

a) Write the overall balanced equation for the cell and calculate the cell emf under standard conditions.

Zn2+ + 2e- Zn E red = -0.76 V

Al3+ + 3e- Al E red = -1.66 V

3Zn2+ + 2Al 3Zn + 2Al3+ Eocell= -0.76 V + 1.66 V = 0.90 V

b) Calculate G° at 298 K.

G = -nFE = -(6 mol e-)

emolVJ96500 0.90 V = -5.2 x105 J or -520 kJ

c) What is the value of the equilibrium constant, K, for this reaction at 298 K? Eo = (RT/nF) ln K; n = 6; K = 1.6 x 1091

Which combination below will undergo a spontaneous reaction?

Half Reaction Table E red Br2(l) + 2e- 2Br–(aq) +1.06 V Cu2+(aq) + 2e- Cu(s) +0.34 V Ni2+(aq) + 2e- Ni(s) −0.28 V Al3+(aq) + 3e- Al(s) −1.66 V

a. Ni 2+(aq) with Br–(aq) b. Cu(s) with Al3+(aq) c. Ni (s) with Br–(aq) d. Br–(aq) with Cu2+(aq) e. Br2(l) with Ni(s)

A voltaic cell is constructed that used the following reaction and operates at 298 K:

2Al(s) + 3Mn2+(aq) 2Al3+(aq) + 3Mn(s) a. What is the EMF of this cell under standard conditions? (Use Appendix D for Eo

red values.)

+1.66 V + -1.18 V = 0.48 V

b. What is the EMF of this cell when [Mn2+] = 0.10 M and [Al3+] = 1.5 M?

E = 0.48 V – (0.0592 V / 6 mol e-) (log (1.5)2/(0.10)3) = 0.45 V 16. Give the symbol, atomic number, and mass number from each of the following:

beta particle : e0

1-01 electron an be toknow now isit sinceor

alpha particle: He nucleus helium a be toknow now isit sinceor 42

42

gamma radiation: 0

0

positron: e01 electron, the toparticle-anti the

neutron: n10

17. Uranium-238, 238U, can undergo beta decay, where two beta particles are released. Write

the equation for this nuclear reaction:

U23892 → 2 0

1 + Pu23894

The same isotope can also undergo alpha decay, releasing one alpha particle.

Write the equation for that mode of decay.

U23892 → 42 + Th234

90

18. Another isotope of uranium, 229U, has a half-life of 58 minutes. Calculate the time, in hours, it would take for only 3.0% of a 100.0g sample to remain.

Since nuclear decay follows first-order kinetics, we can use the same formulas: Instead of concentrations, we can also use mass or the number of disintegrations (a measure of radioactivity). Just depends on the information given.

kln2t1/2 kt

][A][Aln

0

t

kln2min 85 1-min 012.0

min 58ln2k

t)min 012.0(.%100%0.3ln 1-

hr 9.4

min 60hr 1min 292 min 292 t

Since it’s 1st order, we can just use % in the equation. For second

order, we would use the actual masses.