þ produce h + (as h 3 o + ) ions in water (the hydronium ion is a hydrogen ion attached to a water...

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Acid and Bases

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Acid and Bases

Acid and Bases

þ Produce H+ (as H3O+) ions in water (the hydronium

ion is a hydrogen ion attached to a water molecule)

þ Taste sour

þ Corrode metals (react to H2 (g))

þ Are electrolytes (conduct electricity)

þ React with bases to form a salt and water

þ pH is less than 7

þ Turns blue litmus paper to red “Blue to Red A-CID”

Some Properties of Acids

Produce OH- ions in water

Taste bitter, chalky

Are electrolytes (conduct electricity)

Feel soapy, slippery

React with acids to form salts and water

pH greater than 7

Turns red litmus paper to blue “Basic Blue”

Some Properties of Bases

HCl Hydrochloric Acid Stomach acid

HNO3 Nitric Acid Jewelry making

H2SO4 Sulfuric Acid Paper making; Car batteries

H3PO4 Phosphoric Acid Preservative in Coca-Cola

NaOH Sodium hydroxide lye

KOH Potassium hydroxide liquid soap

Ba(OH)2 Barium hydroxide stabilizer for plastics

Mg(OH)2 Magnesium hydroxide “MOM” Milk of magnesia

Al(OH)3 Aluminum hydroxide Maalox (antacid)

Some Common Acids & Bases- (video)

Definition #1: Arrhenius (traditional)

Acids – produce H+ ions (or hydronium ions H3O+)

Bases – produce OH- ions

(problem: some bases don’t have hydroxide ions!)

Acid/Base definitions

Arrhenius acid is a substance that produces H+ (H3O+) in water

Arrhenius base is a substance that produces OH- in water

Strong acid- ionizes completely in aqueous solutionStrong electrolytesHCl, HNO3

Weak acid- releases few hydrogen ions in aqueous solutionHCN and acetic acid (-COOH)

Strong base- ionizes completely in aqueous solutionStrong electrolytes

Weak base- releases few hydroxide ions in aqueous solution

Arrhenius Acids and Bases- (video)

HNO3, HCl, H2SO4 and HClO4 are among the only strong acids.

Strong and Weak Acids/Bases

The strength of an acid (or base) is determined by the amount of IONIZATION. STRONG=100% IONIZED

The strength of an acid (or base) is determined by the amount of IONIZATION. STRONG=100% IONIZED

Weak acids are much less than 100% ionized in water.

One of the best known is acetic acid = CH3COOH

Strong and Weak Acids/Bases

Strong and Weak Acids/Bases

Strong Base: 100% dissociated in water.

NaOH (aq) ---> Na+ (aq) + OH- (aq)

Strong and Weak Acids/Bases

Other common strong bases include KOH and Ca(OH)2.

CaO (lime) + H2O -->

Ca(OH)2 (slaked lime)CaO

Weak base: less than 100% ionized in water

One of the best known weak bases is ammonia

NH3 (aq) + H2O (l) NH4+ (aq) + OH-

(aq)

Strong and Weak Acids/Bases

Strong and Weak Acids/Bases

Acid/Base Definitions- (video)Definition #2: Brønsted

– Lowry

Acids – molecule or ion that is a proton donor

Bases – molecule or ion that is a proton acceptor

A “proton” is really just a hydrogen atom that has lost it’s electron!

A Brønsted-Lowry acid is a proton donorA Brønsted-Lowry base is a proton acceptor

acidconjugate

basebase conjugate

acid

ACID-BASE THEORIESACID-BASE THEORIESThe Brønsted definition means

NH3 is a BASE in water — and water is itself an ACID

BaseAcidAcidBaseNH4

+ + OH-NH3 + H2OBaseAcidAcidBase

NH4+ + OH-NH3 + H2O

Conjugate Pairs- (video)

Acids & Base Definitions

Lewis acid - a substance that accepts an electron pair

Lewis base - a substance that donates an electron pair

Definition #3 – Lewis Definition #3 – Lewis

Formation of hydronium ion is also an excellent example.

Lewis Acids & Bases

• Electron pair of the new O-H bond originates on the Lewis base.

HH

H

BASE

••••••

O—HO—H

H+

ACID

SUMMARY

NO, NO2, CO2, SO2, and SO3 gases from industrial processes can dissolve in atmospheric water to produce acidic solutions.

Burning of fossil fuels by coal-burning power plants, factories, and automobiles

Very acidic rain is known as acid rain.

Acid rain can erode statues and affect

ecosystems.

Acid Rain

example: SO3(g) + H

2O(l) H

2SO

4(aq)

Acid Rain

Amphoteric Compounds

Any species that can react as either an acid or a base is described as amphoteric.

example: waterwater can act as a base

H2SO

4(aq) + H

2O(l) H

3O(aq) + HSO

4Ğ(aq)

NH

3(g) + H

2O(l) NH

4(aq) OHĞ(aq)

acid1 base2 acid2 base1

water can act as an acid

base1 acid2 acid1 base2

More About WaterMore About WaterH2O can function as both an ACID and a

BASE.

In pure water there can be AUTOIONIZATION

Equilibrium constant for water = Kw

Kw = [H3O+] [OH-] = 1.00 x 10-14 at 25 oC

In the self-ionization of water, two water molecules produce a hydronium ion and a hydroxide ion by transfer of a proton.

Self-ionization or Autoionization of Water

H

2O(l) + H

2O(l) H

3O(aq) + OHĞ(aq)

-

Relationship of [H3O+] to [OH–]

Kw = ionization constant of water

Kw = [H3O+] [OH-] = 1.00 x 10-14 at 25 oC

In a neutral solution [H3O+] = [OH-]

so Kw = [H3O+]2 = [OH-]2

and so [H3O+] = [OH-] = 1.00 x 10-7 M

More About WaterMore About WaterOH-

H3O+

OH-

H3O+

Autoionization

Kw = [H+1][OH-1] = 1E-14

Used to find unknown [H+1] or [OH-1] if the other is known or given

If [H+1] = .0003M, find the [OH-1]

1E-14 = [.0003M] [OH-1]

[OH-1]=3.33E-11M

Kw: the Constant of Water or How to find the missing #

The pH scale is a way of expressing the strength of acids and bases. Instead of using very small numbers, we just use pH

Base 10 logarithms (log)The log of a number is the power to which 10

must be raised to get that numberlog 1000 = 3 because 103 = 1000log 0.1 = -1 because 10-1 = 0.1Use your calculator to solve

log (1.25*10-3) = -2.90-log (3.64*10-12) = 11.44

Logarithms

pH = - log [H+]Example: If [H+] =1E-10

pH = - log( 1E-10)pH = - (- 10)pH = 10

Example: If [H+] = 1.8E-5pH = - log 1.8E-5pH = - (- 4.74)pH = 4.74

Calculating the pH

pH Values as Specified [H3O+]

Find the pH of these:1) A 0.15 M solution of Hydrochloric

acid (answer: .8239)2) A 3.00 X 10-7 M solution of Nitric

acid (answer: 6.5229)

STOP

Try These!

If the pH of Coke is 3.12, [H+] = ???

Because pH = - log [H+] then

- pH = log [H+]

Take antilog (10x) of both sides and get

10-pH = [H+][H+] = 10-3.12 = 7.6 x 10-4 M *** to find antilog on your calculator, look for “Shift” or “2nd

function” and then the log button

pH calculations – Solving for H+pH calculations – Solving for H+

A solution has a pH of 8.5. What is the Molarity of hydrogen ions in the solution?

pH calculations – Solving for H+

pH = - log [H+]

8.5 = - log [H+]

-8.5 = log [H+]

Antilog -8.5 = antilog (log [H+])

10-8.5 = [H+]

3.16E-9 M = [H+]

pH = - log [H+]

8.5 = - log [H+]

-8.5 = log [H+]

Antilog -8.5 = antilog (log [H+])

10-8.5 = [H+]

3.16E-9 M = [H+]

Using Logarithms in pH Calculations

Since acids and bases are opposites, pH and pOH are opposites!

pOH does not really exist, but it is useful for changing bases to pH.

pOH looks at the perspective of a base

pOH = - log [OH-]Since pH and pOH are on

opposite ends,pH + pOH = 14

pOH

[OH-]

[H+] pOH

pH

10 -pOH

10 -pH-Log[H+]

-Log[OH

-]

14 -

pOH

14 -

pH

1.0

x 10-1

4

[OH

- ]

1.0

x 10-1

4

[H

+ ]

The pH of rainwater collected in a certain region of the northeastern United States on a particular day was 4.82. What is the H+ ion concentration of the rainwater? (answer: 1.51E-5M)

The OH- ion concentration of a blood sample is 2.5 x 10-7 M. What is the pH of the blood?(answer: 7.3979)

Calculating [H3O+], pH, [OH-], and pOH

Problem 1: Calculate the [H3O+], pH, [OH-], and pOH of a solution of 0.0024 M hydrochloric acid at 25°C.

[H3O+]= 0.0024 M

pH= 2.62

[OH-]= 4.17 E^-12M

pOH= 11.38

End

Problem 2: What is the [H3O+], [OH-], and pOH of a solution with pH = 3.67? Is this an acid, base, or neutral?

[H3O+]= 10^ -3.67 = 2.14E-4 M

[OH-]= 10^-14 / (2.14E-4) = 4.67E-11 M

pOH= 14 – 3.67 = 10.33

Acid

pH meterTests the voltage of the

electrolyteVoltage changes as

hydronium ion concentration changes

Converts the voltage to pHVery cheap, accurateMust be calibrated with a

buffer solution

pH indicatorsIndicators are compounds that will

change color in the presence of an acid or base.

Indicators are either weak acids or weak bases

Indicators only work in a specific range of pH

Some dyes are natural, like radish skin or red cabbage

Neutralization ReactionsStrong Acid-Strong Base NeutralizationIn aqueous solutions, neutralization is the

reaction of hydronium ions and hydroxide ions to form water molecules.

H3O+(aq) + OH(aq) 2H2O(l)

A salt is an ionic compound composed of a cation from a base and an anion from an acid.

HCl(aq) + NaOH(aq) NaCl(aq) H2O(l)

Titration is the controlled addition and measurement of the amount of a solution of known concentration required to react completely with a measured amount of a solution of unknown concentration.

The point at which the two solutions used in a titration are present in chemically equivalent amounts is the equivalence point.

The point in a titration at which an indicator changes color is called the end point of the indicator.

Titration

Setup for titrating an acid with a base

Titration Curve for a Strong Acid and a Strong Base

Titration Curve for a Weak Acid and a Strong Base

TitrationTitration1. Add solution from the buret

to the flask.2. Reagent (base) reacts with

compound (acid) in solution in the flask.

3. Indicator shows when exact stoichiometric reaction has occurred. (Acid = Base)

This is called the END POINT where NEUTRALIZATION has occurred.

Titration Math(short version)

M1 • V1 = M2 • V2

Moles H3O+ = Moles OH-

Standard Solution = NaOH

BURET

Solution of Unknown Concentration = HCl

Titration using a pH MeterSet up the buret and the chemicals the same except no

Phenolphthalein is addedSet up the Nova with a pH sensor attached and insert the

probe into the Unknown solution. Proceed with the addition of the base solution until the graph looks like the one attached

The volume where the large

jump takes place is the End

Point, moles Base=moles AcidBegin calculations

Molarity and Titration

1. Start with the balanced equation for the neutralization reaction, and determine the chemically equivalent amounts of the acid and base.

2. Determine the moles of acid (or base) from the known solution used during the titration.

3. Determine the moles of solute of the unknown solution used during the titration using the balanced equation.

4. Determine the molarity of the unknown solution.

Problem: Determine the molarity of an acidic solution, 10 mL HCl, by titration.

(HCl of unknown molarity in the flask, 5.0 x 10-3 M NaOH in the buret)

Titrate the acid with a standard base solution 20.00 mL of 5.0 × 10−3 M NaOH was titrated

1. Write the balanced neutralization reaction equation. HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)

1 mol HCL reacts with 1 mol NaOH

Determine the chemically equivalent amounts of HCl and NaOH.

3. Calculate the number of moles of NaOH used in the titration. Then calculate the number of moles of HCl initially in the flask.

20.0 mL of 5.0 × 10−3 M NaOH is needed to reach the end point

-3-45.0 10 mol NaOH 1 L

20 mL 1.0 10 mol NaOH used1 L 1000 mL

-4-21.0 10 mol HCl 1000 mL

1.0 10 M HCl10.0 mL 1 L

Amount of HCl = mol NaOH = 1.0 × 10−4 mol

4. Calculate the molarity of the HCl solution

Sample Problem FIn a titration, 27.4 mL of 0.0154 M Ba(OH)2

is added to a 20.0 mL sample of HCl solution of unknown concentration until the equivalence point is reached. What is the molarity of the acid solution?

Sample Problem F SolutionGiven: volume and concentration of known solution

= 27.4 mL of 0.0154 M Ba(OH)2

Unknown: molarity of acid solution

Solution:

1. balanced neutralization equation chemically equivalent amounts

Ba(OH)2 + 2HCl BaCl2 + 2H2O1 mol 2 mol 1 mol 2 mol

Sample Problem F Solution, continued

2. volume of known basic solution used (mL) amount of base used (mol)

mol Ba(OH)2

1 LmL of Ba(OH)

2 solution

1 L

1000 mLmol Ba(OH)

2

3. mole ratio, moles of base used moles of acid used from unknown solution

2 mol HCl

mol Ba(OH)2

mol of Ba(OH)2 in known solution mol HCl

Sample Problem F Solution, continued

4. volume of unknown, moles of solute in unknown molarity of unknown

amount of solute in unknown solution (mol)

volume of unknown solution (mL)

1000 mL

1 L

molarity of unknown solution

Sample Problem F Solution, continued

1. 1 mol Ba(OH)2 for every 2 mol HCl.

2.

3.

0.0154 mol Ba(OH)2

1 L 24.7 mL of Ba(OH)

2 solution

1 L

1000 mL4.22 10-4mol Ba(OH)

2

2 mol HCl

1 mol Ba(OH)2

4.22 10Ğ4 mol of Ba(OH)2

8.44 10Ğ4 mol HCl

3.80

3.80

7.61-4

-4

Sample Problem F Solution, continued4.

8.44 10-4 mol HCl

20.0 mL

1000 mL

1 L 4.22 10-2 M HCl

End

3.87.61