Titrations of acids and bases

HA + H 2 O H 3 O + + A -

Titrations of acids and bases HA + H 2 O H 3 O + + A - B + H 2 O OH - + HB +

Titrations of acids and bases HA + H 2 O H 3 O + + A - B + H 2 O OH - + HB + H 3 O + + OH - 2 H 2 O

At equivalence point [H 3 O + ] = [OH - ] For strong acid – strong base

H 3 O + + OH - 2 H 2 O At equivalence point [H 3 O + ] = [OH - ] K w = [ H 3 O + ] = [OH - ] =

H 3 O + + OH - 2 H 2 O At equivalence point [H 3 O + ] = [OH - ] K w = [ H 3 O + ] = [OH - ] = pH = 7

Titrations of acids and bases HA + H 2 O H 3 O + + A - B + H 2 O OH - + HB + H 3 O + + OH - 2 H 2 O

Volume of base added Start with acid solution pH 7

4 10

100 ml 0.1 M HCl

Titrate with 0.1 M NaOH

100 ml 0.1 M HCl Titrate with 0.1 M NaOH H 3 O + + OH - 2 H 2 O

100 ml 0.1 M HCl Titrate with 0.1 M NaOH H 3 O + + OH - 2 H 2 O Every part of a mole of NaOH added reduces the moles of H 3 O + by an equal amount.

100 ml 0.1 M HCl Titrate with 0.1 M NaOH H 3 O + + OH - 2 H 2 O As NaOH solution is added, the overall volume increases. This further decreases [H 3 O + ].

100 ml 0.1 M HCl Titrate with 0.1 M NaOH Neutralization and dilution; both increase the pH of the solution.

100 ml 0.1 M HCl Titrate with 0.1 M NaOH Add 50 ml of NaOH solution.

100 ml 0.1 M HCl Titrate with 0.1 M NaOH Add 50 ml of NaOH solution. Start with 0.1 L x 0.1 M H 3 O +

100 ml 0.1 M HCl Titrate with 0.1 M NaOH Add 50 ml of NaOH solution. Start with 0.1 L x 0.1 M H 3 O moles H 3 O +

100 ml 0.1 M HCl Titrate with 0.1 M NaOH Add 50 ml of NaOH solution moles H 3 O + Add 0.05 L x 0.1 M OH -

100 ml 0.1 M HCl Titrate with 0.1 M NaOH Add 50 ml of NaOH solution moles H 3 O + Add 0.05 L x 0.1 M OH moles OH -

100 ml 0.1 M HCl Titrate with 0.1 M NaOH Add 50 ml of NaOH solution moles H 3 O moles OH - H 3 O + is neutralized by OH - 1:1 ratio

100 ml 0.1 M HCl Titrate with 0.1 M NaOH Add 50 ml of NaOH solution moles H 3 O moles OH - Remaining H 3 O + = 0.01 moles – moles

100 ml 0.1 M HCl Titrate with 0.1 M NaOH Add 50 ml of NaOH solution moles H 3 O moles OH - Remaining H 3 O + = moles Total volume = 150 ml = 0.15 L

100 ml 0.1 M HCl Titrate with 0.1 M NaOH Add 50 ml of NaOH solution moles H 3 O moles OH - [H 3 O + ]= moles 0.15 L

100 ml 0.1 M HCl Titrate with 0.1 M NaOH Add 50 ml of NaOH solution moles H 3 O moles OH - [H 3 O + ]= moles 0.15 L = M

100 ml 0.1 M HCl Titrate with 0.1 M NaOH Add 50 ml of NaOH solution moles H 3 O moles OH - [H 3 O + ]= moles 0.15 L = M pH = 1.48

100 ml 0.1 M HCl Titrate with 0.1 M NaOH Add 99.9 ml of NaOH solution.

100 ml 0.1 M HCl Titrate with 0.1 M NaOH Add 99.9 ml of NaOH solution moles H 3 O +

100 ml 0.1 M HCl Titrate with 0.1 M NaOH Add 99.9 ml of NaOH solution moles H 3 O L x 0.1 M = moles OH -

100 ml 0.1 M HCl Titrate with 0.1 M NaOH Add 99.9 ml of NaOH solution moles H 3 O L x 0.1 M = moles OH - Moles H 3 O + = 0.01 – = 1 x 10 -5

100 ml 0.1 M HCl Titrate with 0.1 M NaOH Add 99.9 ml of NaOH solution moles H 3 O L x 0.1 M = moles OH - Moles H 3 O + = 0.01 – = 1 x Total volume = 0.1 L L = L

100 ml 0.1 M HCl Titrate with 0.1 M NaOH Add 99.9 ml of NaOH solution moles H 3 O L x 0.1 M = moles OH - [H 3 O + ]= 1 x moles L

100 ml 0.1 M HCl Titrate with 0.1 M NaOH Add 99.9 ml of NaOH solution moles H 3 O L x 0.1 M = moles OH - [H 3 O + ]= 1 x moles L = 5 x 10 -5

100 ml 0.1 M HCl Titrate with 0.1 M NaOH Add 99.9 ml of NaOH solution moles H 3 O L x 0.1 M = moles OH - [H 3 O + ]= 1 x moles L = 5 x pH = 4.3

4.3

+ 0.1 ml will give pH = 7

If more 0.1 M NaOH solution is added after the equivalence point, there is no H 3 O + to neutralize it.

At pH = 7 the volume is 0.2 L

Add 0.1 mL 0.1 M NaOH

At pH = 7 the volume is 0.2 L Add 0.1 mL 0.1 M NaOH Add L x 0.1 M = moles OH - [OH - ] = 1 x moles L = 5 x M

At pH = 7 the volume is 0.2 L Add 0.1 mL 0.1 M NaOH Add L x 0.1 M = moles OH - [OH - ] = 1 x moles L = 5 x M pOH = 4.3 pH = 9.7

4.3 pH = 7 V = ml V = 200 ml pH = 9.7 V = ml

Titrating a weak acid with a strong base

HA + H 2 O H 3 O + + A - K a < 1

Titrating a weak acid with a strong base HA + H 2 O H 3 O + + A - K a < 1 Much less than 100% dissociation.

Titrating a weak acid with a strong base HA + H 2 O H 3 O + + A - K a < 1 Much less than 100% dissociation. Every OH - added neutralizes an H 3 O + and shifts the equilibrium to the right.

Titrate 100 ml 0.10 M CH 3 COOH With 0.10 M NaOH

Titrate 100 ml 0.10 M CH 3 COOH With 0.10 M NaOH K a = 1.8 x 10 -5

Titrate 100 ml 0.10 M CH 3 COOH With 0.10 M NaOH K a = 1.8 x = [H 3 O + ][CH 3 COO - ] [CH 3 COOH] 0 ml NaOH

Titrate 100 ml 0.10 M CH 3 COOH With 0.10 M NaOH K a = 1.8 x = [H 3 O + ][CH 3 COO - ] [CH 3 COOH] 0 ml NaOH pH = 2.87

Titrate 100 ml 0.10 M CH 3 COOH With 0.10 M NaOH K a = 1.8 x = [H 3 O + ][CH 3 COO - ] [CH 3 COOH] 50 ml NaOH.1 L x 0.1 M = 0.01 moles CH 3 COOH 0.05L x 0.1 M = moles NaOH

CH 3 COOH + H 2 O H 3 O + + CH 3 COO - + OH - 2 H 2 O

CH 3 COOH + H 2 O H 3 O + + CH 3 COO - + OH - 2 H 2 O Net result: for every OH - added, there is one less CH 3 COOH and one more CH 3 COO -

Titrate 100 ml 0.10 M CH 3 COOH With 0.10 M NaOH K a = 1.8 x = [H 3 O + ][CH 3 COO - ] [CH 3 COOH] 50 ml NaOH.1 L x 0.1 M = 0.01 moles CH 3 COOH 0.05L x 0.1 M = moles NaOH moles CH 3 COOH = moles

Titrate 100 ml 0.10 M CH 3 COOH With 0.10 M NaOH K a = 1.8 x = [H 3 O + ][CH 3 COO - ] [CH 3 COOH] 50 ml NaOH.1 L x 0.1 M = 0.01 moles CH 3 COOH 0.05L x 0.1 M = moles NaOH moles CH 3 COOH = moles moles CH 3 COO -

Titrate 100 ml 0.10 M CH 3 COOH With 0.10 M NaOH K a = 1.8 x = [H 3 O + ][CH 3 COO - ] [CH 3 COOH] 50 ml NaOH moles CH 3 COOH = moles moles CH 3 COO L solution

[CH 3 COOH] = moles moles CH 3 COO L = M

[CH 3 COOH] = moles moles CH 3 COO L = M K a = 1.8 x = [H 3 O + ][CH 3 COO - ] [CH 3 COOH] Buffer solution: pK a = pH = 4.74

Titrate 100 ml 0.10 M CH 3 COOH With 0.10 M NaOH K a = 1.8 x = [H 3 O + ][CH 3 COO - ] [CH 3 COOH] 100 ml NaOH 0.01 moles CH 3 COO L solution

Titrate 100 ml 0.10 M CH 3 COOH With 0.10 M NaOH K a = 1.8 x = [H 3 O + ][CH 3 COO - ] [CH 3 COOH] 100 ml NaOH 0.01 moles CH 3 COO L solution [CH 3 COO - ] = 0.05 M

Titrate 100 ml 0.10 M CH 3 COOH K a = 1.8 x = [H 3 O + ][CH 3 COO - ] [CH 3 COOH] [CH 3 COO - ] = 0.05 M K b = KwKw KaKa = x = 5.6 x K b = [CH 3 COOH][OH - ] [CH 3 COO - ]

Titrate 100 ml 0.10 M CH 3 COOH K a = 1.8 x = [H 3 O + ][CH 3 COO - ] [CH 3 COOH] [CH 3 COO - ] = 0.05 M K b = KwKw KaKa = x = 5.6 x K b = [CH 3 COOH][OH - ] [CH 3 COO - ] = y2y2 0.05

Titrate 100 ml 0.10 M CH 3 COOH K a = 1.8 x = [H 3 O + ][CH 3 COO - ] [CH 3 COOH] K b = KwKw KaKa = x = 5.6 x K b = [CH 3 COOH][OH - ] [CH 3 COO - ] = y2y y 2 = (5.6 x )(0.05) = 2.8 x

Titrate 100 ml 0.10 M CH 3 COOH K a = 1.8 x = [H 3 O + ][CH 3 COO - ] [CH 3 COOH] K b = KwKw KaKa = x = 5.6 x K b = [CH 3 COOH][OH - ] [CH 3 COO - ] = y2y y 2 = (5.6 x )(0.05) = 2.8 x y = 5.3 x 10 -6

Titrate 100 ml 0.10 M CH 3 COOH K a = 1.8 x = [H 3 O + ][CH 3 COO - ] [CH 3 COOH] K b = KwKw KaKa = x = 5.6 x K b = [CH 3 COOH][OH - ] [CH 3 COO - ] = y2y y 2 = (5.6 x )(0.05) = 2.8 x y = 5.3 x pOH =5.3; pH = 8.7

After the equivalence point, OH - is being added to a saturated buffer system.

After the equivalence point, OH - is being added to a saturated buffer system. pH increases rapidly