A25.0-ml solution of 0.100 m ch3cooh
istitrated with a 0.200 m koh solution. calculate theph
after the following additions of the koh solution : (a) 0.0 ml,(b)
5.0 ml, (c) 10.0 ml, (d) 12.5 ml, (e) 15.0 ml.

a) pH = 2,88

b) pH = 4,58

c) pH = 5,36

d) pH = 8,79

e) pH = 12,10

Explanation:

In a titration of a strong base (KOH) with a weak acid (CH₃COOH) the reaction is:

CH₃COOH + KOH → CH₃COOK + H₂O

a) Here you have just CH₃COOH, thus:

CH₃COOH ⇄ CH₃COO⁻ + H⁺ where ka =1,74x10⁻⁵ and pka = 4,76

When this reaction is in equilibrium:

[CH₃COOH] = 0,100 -x

[CH₃COO⁻] = x

[H⁺] = x

Thus, equilibrium equation is:

1,74x10⁻⁵ =

The equation you will obtain is:

x² + 1,74x10⁻⁵x - 1,74x10⁻⁶ = 0

Solving:

x = -0,0013278193 ⇒ No physical sense. There are not negative concentrations

x = 0,0013104193

As x = [H⁺] and pH = - log [H⁺]

pH = 2,88

b) Here, it is possible to use:

CH₃COOH + KOH → CH₃COOK + H₂O

With adition of 5,0 mL of 0,200M KOH solution the initial moles are:

CH₃COOH = = 2,5x10⁻³ mol

KOH = = 1,0x10⁻³ mol

CH₃COOK = 0.

In equilibrium:

CH₃COOH = 2,5x10⁻³ mol - 1,0x10⁻³ mol = 1,5x10⁻³ mol

KOH = 0 mol

CH₃COOK = 1,0x10⁻³ mol

Now, you can use Henderson–Hasselbalch equation:

pH = 4,76 + log

pH = 4,58

c) With adition of 10,0 mL of 0,200M KOH solution the initial moles are:

CH₃COOH = = 2,5x10⁻³ mol

KOH = = 2,0x10⁻³ mol

CH₃COOK = 0.

In equilibrium:

CH₃COOH = 2,5x10⁻³ mol - 2,0x10⁻³ mol = 0,5x10⁻³ mol

KOH = 0 mol

CH₃COOK = 2,0x10⁻³ mol

Now, you can use Henderson–Hasselbalch equation:

pH = 4,76 + log

pH = 5,36

d) With adition of 12,5 mL of 0,200M KOH solution the initial moles are:

CH₃COOH = = 2,5x10⁻³ mol

KOH = = 2,5x10⁻³ mol

CH₃COOK = 0.

Here we have the equivalence point of the titration, thus, the equilibrium is:

CH₃COO⁻ + H₂O ⇄ CH₃COOH + OH⁻ kb = kw/ka where kw is equilibrium constant of water = 1,0x10⁻¹⁴; kb = 5,75x10⁻¹⁰

Concentrations is equilibrium are:

[CH₃COOH] = x

[CH₃COO⁻] = 0,06667-x

[OH⁻] = x

Thus, equilibrium equation is:

5,75x10⁻¹⁰ =

The equation you will obtain is:

x² + 5,75x10⁻¹⁰x - 3,83x10⁻¹¹ = 0

Solving:

x = -0.000006188987⇒ No physical sense. There are not negative concentrations

x = 0.000006188

As x = [OH⁻] and pOH = - log [OH⁻]; pH = 14 - pOH

pOH = 5,21

pH = 8,79

e) The excess volume of KOH will determine pH:

With 12,5mL is equivalence point, the excess volume is 15,0 -12,5 = 2,5 mL

2,5x10⁻³ L × ÷ 0,040 L = 0,0125 = [OH⁻]

pOH = - log [OH⁻]; pH = 14 - pOH

pOH = 1,90

pH = 12,10

I hope it helps!

the volume increases