Question

what ratio of nacn to hcn is needed to prepare a ph 10.10 buffer? (kₐ of hcn is 4.9 × 10⁻¹⁰)

Explanation:

Step1: Use Henderson - Hasselbalch equation

The Henderson - Hasselbalch equation is $pH = pK_{a}+\log\frac{[A^{-}]}{[HA]}$. Here, $A^{-}$ is $CN^{-}$ (from $NaCN$) and $HA$ is $HCN$. First, calculate $pK_{a}$: $pK_{a}=-\log(K_{a})$. Given $K_{a} = 4.9\times10^{-10}$, so $pK_{a}=-\log(4.9\times 10^{-10})=\log(10^{10})-\log(4.9)=10 - 0.69 = 9.31$.

Step2: Substitute values into Henderson - Hasselbalch equation

We know $pH = 10.10$ and $pK_{a}=9.31$. Substitute into $pH = pK_{a}+\log\frac{[CN^{-}]}{[HCN]}$. Then $10.10=9.31+\log\frac{[CN^{-}]}{[HCN]}$. Rearrange to get $\log\frac{[CN^{-}]}{[HCN]}=10.10 - 9.31 = 0.79$.

Step3: Solve for the ratio

If $\log\frac{[CN^{-}]}{[HCN]} = 0.79$, then $\frac{[CN^{-}]}{[HCN]}=10^{0.79}$. Using a calculator, $10^{0.79}\approx6.2$.

Answer:

The ratio of $[NaCN]$ to $[HCN]$ is approximately $6.2$.