Question

Hydrogen cyanide gas (HCN), a powerful respiratory inhibitor, is highly toxic. It is a very weak acid (Ka = 6.2 x 10-10) when dissolved in water. If a 50.0 ml sample of 0.100 M HCN is titrated with 0.100 M NaOH, calculate the pH of the solution After 8.00 mL of 0.100 M NaOH has been added At the halfway point of the titration At the equivalence point of the titration

Answer 1

Start with writting acid-base system in question is

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

I would like to divide calculations in three steps-
1.. mmol of added base = (8.00 mL)(0.100 mmol/mL) = 0.800 mmol

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

The added base will convert 0.800 mmol of HCN into CN⁻,
therefore [H₃O⁺] = Kₐ[HCN]/[CN⁻] = (6.2 × 10⁻¹⁰)(4.200 mmol/58.0 mL)/(0.800 mmol/58 mL) = 1.18 × 10⁻¹⁰ ⇒ pH = –log[H₃O⁺] = 9.93

Note that the volumes canceled out - all we needed was the HCN to CN⁻ mole ratio.

2.. at the halfway point of the titration, [HCN] = [CN⁻], so [H₃O⁺] = Kₐ ⇒ pH = pKₐ = 9.21

3..At the equivalence point, the solution has been converted into a NaCN solution; 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. The easiest way to think about the solution at the equivalence point is to write the equilibrium involving the conjugate base:

CN⁻ + H₂O ⇌ HCN + OH⁻ and Kb(CN⁻) = Kw/Kₐ(HCN) = 10⁻¹⁴/(6.2 × 10⁻¹⁰) = 1.61 × 10⁻⁵ = [HCN][OH⁻]/[CN⁻] ⇒ now [OH⁻] = [HCN] to a good approximation, so 1.61 × 10⁻⁵ = [OH⁻]²/(0.05) where I've assumed that [CN⁻] >> [OH⁻] - an assumption we can check:

Solving that expression, [OH⁻] = 8.98 × 10⁻⁴ which means the assumption is OK, but off by almost 2%. (You can use a quadratic formula to do better.) Anyway pOH ≅ 3.05 so pH = 10.95