Question

What is the pH at the equivalence point in the titration of 100 mL of 0.10 M HCN (CN- Kb = 2.0 × 10–5) with 0.10 M NaOH (100 mL)?

Answer 1

Calculate moles of acid present:

(0.10 mol/L) (0.1 L) = 0.01 moles

HCN + NaOH ---> NaCN + H₂O

There is a 1:1 molar ratio between acid and sodium hydroxide.

Total volume is 100 mL +100 mL=200 mL=0.2 mL

Calculate molarity of Sodium cyanide:

0.01 mol / 0.2 L = 0.05 M

2) Calculate the Ka of CN:

K_w = K_aK_b

1.00 x 10^-14 = (2 x 10^-5 ) (x)

x = 5. x 10^-10

3) Calculate pH of the solution:

5. x 10^-10 = [(x) (x)] / 0.05

x =0.0001 M (this is the hydroxide ion concentration)

pOH = -log 0.0001=4

pH = 14-4=10

Therefore, the pH at the equivalence point is 10