Question

HCN gas, a powerful respiratory inhibitor, is highly toxic. It is a very weak acid (K_a = 6.2 x 10^-10) when dissolved in water. If a 50.00-mL sample of 0.100 M HCN is titrated with 0.100 M NaOH, calculate the pH of the solution:

a. before the titration starts.

b. after 8.00 mL of 0.100 M NaOH has been added.

c. at the half-way point of the titration.

d. at the equivalence point of the titration.

e. after 52.00 mL of 0.100 M NaOH has been added.

Answer 1

a.      

The acid-base system in question is

HCN + H₂O ⇌ CN⁻ + H₃O⁺ Kₐ = [CN⁻][H₃O⁺]/[HCN] = 6.2 × 10⁻¹

here a 50.00-mL sample of 0.100 M HCN and K_a = 6.2 x 10^-10

for week acid Ka = x2/M-x

6.2 x 10^-10 = x2 /0.100-x

0.100 –x ≈0.100

x = 7.9*10^-6

[H3O+]=[CN-] = 7.9*10^-6

pH = - log [H3O+]

pH = - log [7.9*10^-6]

pH = 5.10

b.

mmol of base NaOH= (8.00 mL)(0.100 mmol/mL) = 0.800 mmol

present mmol of HCN = (50.00 mL)(0.100 mmol/mL) = 5.000 mmol

at this point the added NaOH will convert 0.800 mmol of HCN into CN⁻,
so [H₃O⁺] in the reaction can be calcualted as follows:

[H₃O⁺] = Kₐ[HCN]/[CN⁻]

= (6.2 × 10⁻¹⁰)(4.200 mmol/58.0 mL)/(0.800 mmol/58 mL)

= 1.18 × 10⁻¹⁰

So pH at this point will be

pH = –log[H₃O⁺]

pH = 9.93

c.

at the halfway point of the titration pH will be equal to pKₐ of acid

Kₐ =6.2 × 10⁻¹⁰

pKₐ = 9.21
so pH = 9.21

d.

at the equivalence point of the titration the solution has been converted into a NaCN solution means salt; moles added base equals moles of original HCN and the volume of the solution is 100 mL.

Therefore [CN⁻] = (5.000 mmol)/(100 mL) = 0.05 M.

CN⁻ + H₂O ⇌ HCN + OH⁻

And Kb(CN⁻) for this reaction can be calcualted as follows:

Kb(CN⁻) = Kw/Kₐ(HCN)

Kb(CN⁻) = 10⁻¹⁴/(6.2 × 10⁻¹⁰)

Kb(CN⁻) = 1.61 × 10⁻⁵

Then;

Kb(CN⁻) = [HCN][OH⁻]/[CN⁻]

now [OH⁻] = [HCN]

so, 1.61 × 10⁻⁵ = [OH⁻]²/(0.05)

here assumed that [CN⁻] >> [OH⁻] -

then;

[OH⁻] = 8.97 × 10⁻⁴

pOH = -log [OH⁻]

pOH =-log 8.97 × 10⁻⁴

pOH = 3.05

pH + pOH = 14

pH = 14-3.05

pH = 10.95