Question

A mixture of NaCN and NaHSO4 consists of a total of 0.60 mol. When the mixture is dissolved in 1.0 L of water and comes to equilibrium the pH is found to be 9.6.

Find the amount of NaCN in the mixture.

Answer 1

Solution.

As the pH of a resulting solution is basic, the predominant form is the products of sodium cyanide hydrolysis. The reaction between acidic hydrosulfite-ions and basic cyanide-ions gives the equivalent amount of undissociated HCN:

The acidity of this solution is described by Henderson's equation:

The pKa for HCN is 9.2, so

Hovewer, [HCN] = n(NaHSO₄), and [CN^-]+[HCN] = n(NaCN), so [CN^-] =  n(NaCN) - n(NaHSO₄).

We can compose a system of equations:

The solution of this system is

n(NaCN) = 0.467 moles;

n(NaHSO₄) = 0.133 moles.