Question

1. In the titration of 25.00mL of 0.10M HCN using 0.10M NaOH as the titrant, what is the pH after the addition of 12.50mL of NaOH?

2. In the titration of 25.00mL of 0.10M HCN using 0.10M NaOH as the titrant, what is the pH after the addition of 25.00mL of NaOH?

3. What is the pH of 0.050M NaCN? Ka for HCN is 4.9x10-10.

Answer 1

1)

nuetrilization reaction between HCN and NaOH is

NaOH + HCN NaCN + H₂O

no. of mole = molarity volume of solution in liter

no. of mole of HCN = 0.10 0.025 = 0.0025 mole

no. of mole of NaOH = 0.10 0.01250 = 0.00125 mole

0.00125 mole of NaOH react with 0.00125 mole of HCN to form salt and water

mole of HCN remain in solution = 0.0025 - 0.00125 = 0.00125 mole

total volume of solution = 25 + 12.50 = 37.50 ml = 0.03750 liiter

molarity = no. of mole / volume of solution in liter

molarity of HCN = 0.00125 / 0.03750 = 0.0333 M

HCN dissociate as

HCN + H₂O H₃O⁺ + CN^-

K_a  = [H₃O⁺] [CN^-] / [HCN]

but [H₃O⁺] = [CN^-] = x

then K_a = [x] [x] / [HCN]

[x]² = K_a [HCN] = 4.9 10^-10 0.0333 = 1.6317 10^-11

[x] = 4.039 10^-6  =  [H₃O⁺] = [CN^-] = x

pH = -log[H₃O⁺] = -log(4.039 10^-6) = 5.39

2)

nuetrilization reaction between HCN and NaOH is

NaOH + HCN NaCN + H₂O

no. of mole = molarity volume of solution in liter

no. of mole of HCN = 0.10 0.025 = 0.0025 mole

no. of mole of NaOH = 0.10 0.025 = 0.0025 mole

0.0025 mole of NaOH react with 0.0025 mole of HCN to form salt and water

mole of HCN remain in solution = 0.00 mole

pH = 7

3)

HCN dissociate as

HCN + H₂O H₃O⁺ + CN^-

K_a  = [H₃O⁺] [CN^-] / [HCN]

but [H₃O⁺] = [CN^-] = x

then K_a = [x] [x] / [HCN]

[x]² = K_a [HCN] = 4.9 10^-10 0.050 = 2.45 10^-11

[x] = 4.95 10^-6  =  [H₃O⁺] = [CN^-] = x

pH = -log[H₃O⁺] = -log(4.95 10^-6) = 5.30