Question

What is the pH of 40.0 mL of a solution that is 0.15 M in CN� and 0.29 M in HCN? For HCN, use Ka = 4.9 * 10^-9

Answer 1

This is a buffer solution because it consists of a weak acid (HCN) and its conjugate base (CN-)
Since it's a buffer solution, the pH is equal to the pKa of the solution, you would use the henderson-hasslebach equation:
pH = pKa - log([acid]/[base])
pH = -log(4.9*10-9) - log(.29/.15)
pH = about 8

Answer 2

A 40.0 mL solution that is 0.15 M in CN⁻  and 0.29 M in HCN has a pH of 8.0.

We have a buffer system formed by 0.29 M HCN (weak acid) and 0.15 M CN⁻ (its conjugate base).

The function of a buffer system is to prevent abrupt changes in the pH when acids or bases are added.

Given the acid ionization constant (Ka), we can calculate the pH of a buffer system using the Henderson-Hasselbach's equation.

$pH = pKa + log\frac{[base]}{[acid]} \\\\pH = -log(4.9 \times 10^{-9} ) + log\frac{0.15}{0.29} = 8.0$

A 40.0 mL solution that is 0.15 M in CN⁻  and 0.29 M in HCN has a pH of 8.0.

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