Question

4. The following shows titration of hydrocyanic acid(HCN) with potassium hydroxide (KOH).

100ml of 0.02M HCN was titrated with 0.050M of KOH. (Ka of HCN is 4.9x10^-10)

HCN_(aq) +KOH_(aq)----->KCN_(aq)+H₂O_(l)

a) How much KOH solution is needed to reach the equilence point?

b)Calculate pH at the equivalence point.

c)Calculate pH when 15mL of KOH was added.

d)Calculate pH when 50mL of KOH was added.

Answer 1

a)

mmol of acid = MV = 100*0.02 = 2

mmol of base required = mmol of acid = 2

Vbase = mmol/Mbase = 2/0.05 = 40 mL of KOH

b)

pH in equivalence point:

Vtotal = 40+100 = 140 mL

[CN-] = mmol/V = 2/140 = 0.01428

expect hydrolysis

since CN- = A- is formed

the next equilibrium is formed, the conjugate acid and water

A- + H2O <-> HA + OH-

The equilibrium s best described by Kb, the base constant

Kb by definition since it is an base:

Kb = [HA ][OH-]/[A-]

get ICE table:

Initially

[OH-] = 0

[HA] = 0

[A-] = M

the Change

[OH-] = + x

[HA] = + x

[A-] = - x

in Equilibrium

[OH-] = 0 + x

[HA] = 0 + x

[A-] = M - x

substitute in Kb expression

Kb = [HA ][OH-]/[A-]

Kb = (10^-14)/(4.9*10^-10) =

2.04*10^-5= x*x/(0.01428-x)

solve for x

x^2 + Kb*x - M*Kb = 0

solve for x with quadratic equation

x = OH- =5.29*10^-4

[OH-]  =5.29*10^-4

pOH = -log(OH-) = -log(5.29*10^-4 = 3.28

pH = 14-3.28= 10.72

pH = 10.72

c)

mmol of acid = MV = 0.02*100 = 2

mmol of base = MV = 0.05*15 = 0.75

pH = pKa + log(CN-/HCN)

pH = -log(4.9*10^-10) + log((0.75)/(2-0.075))

pH = 8.90

d)

mmol of acid = MV = 0.02*100 = 2

mmol of base = MV = 0.05*50= 2.5

mmol of OH- left = 0.5

Vtotal = 50+100 = 150

[Oh-]= 0.5/150 = 0.00333

pH = 14 + log(0.00333) = 11.52