Question

25.0 mL of a 0.100 M HCN solution is titrated with 0.100 M KOH. Ka of HCN=6.2 x 10^-10.
a)What is the pH when 25mL of KOH is added (this is the equivalent point).
b) what is the pH when 35 mL of KOH is added.

Answer 1

a) When 25.0mL of KOH is added

Molarity of HCN = 0.100M

Volume of HCN = 25.0 mL

No. of mol of HCN is

Molarity of KOH = 0.100M

When 25.0 mL of KOH is added, the no. of mol of HCN and that of KOH is the same.

So, the no. of mol of CN^- formed will also be equal to the no. of mol of HCN and KOH.

Therefore, no. of mol of CN^- = 2.50 x 10^-3 mol

Total volume = 25.0 mL + 25.0 mL = 50.0 mL = 50.0 x 10^-3L

Concentration of CN^- is

The CN^- will undergo hydrolysis

The ICE table is

	CN-	H2O	HCN	OH-
I	0.0500M	-	0	0
C	-x	-	+x	+x
E	0.0500-x	-	x	x

The K_b of CN^- 1.61 x 10^-5 is very less, so the value of x will be very less such that it can be neglected.

Check if the approximation is valid or not.

If the % of [OH^-] with respect to initial [CN^-] is less than 5%, then the approximation of x is valid.

% whic is less than 5%

So, the approximation holds good.

Now, find pOH from [OH^-]

Therefore, pH at equivalence point is 10.95

b) When 35.0mL of KOH is added.

Molarity of KOH = 0.100M

No. of mol of OH^- taken is

No. of mol of HCN = 2.50 x 10^-3mol

Excess no. of mol of OH^- is

Total volume = 25.0 mL + 35.0 mL = 60.0 mL = 60.0 x 10^-3L

Therefore, concentration of excess OH^- is

Therefore, pH of the solution when 35.0 mL of KOH is added is 12.22