What ratio of NaCN to HCN is needed to prepare a pH 9.80 buffer? (Ka of HCN is 4.9 × 10⁻¹⁰)

Answer 1

Use the Henderson-Hasselbalch equation to solve this problem.

pH = pKa + log([A^-]/[HA]), where [A^-]/[HA] would represent the ratio [CN^-]/[HCN]. Remember that the source of CN^- is NaCN.

The relationship between Ka and pKa is the same as the relationship between [H⁺] and pH; that is

pKa = -log(Ka). The pKa of HCN is -log(4.9 × 10^-10) = 9.31.

Now, we can solve for [CN^-]/[HCN]

9.80 = 9.31 + log( [CN^-]/[HCN] )

0.49 = log( [CN^-]/[HCN] )

[CN^-]/[HCN] = 10^0.49 = 3.1