Question

what is the ph of a buffer made from 0.130 mol of hcno (kₐ = 3.5 × 10⁻⁴) and 0.410 mol of nacno in 2.0 l of solution? answer:

Explanation:

Step1: Use Henderson - Hasselbalch equation

The Henderson - Hasselbalch equation is \(pH = pK_{a}+\log\frac{[A^{-}]}{[HA]}\). First, calculate \(pK_{a}\) using the formula \(pK_{a}=-\log K_{a}\). Given \(K_{a} = 3.5\times10^{-4}\), then \(pK_{a}=-\log(3.5\times 10^{-4})\).

$$
LATEXBLOCK0
$$

The concentration of \(HA\) ( \(HCN\) ) is \(c_{HA}=\frac{n_{HA}}{V}\), where \(n_{HA} = 0.130\space mol\) and \(V = 2.0\space L\), so \(c_{HA}=\frac{0.130}{2.0}=0.065\space M\). The concentration of \(A^{-}\) ( \(CN{O}^{-}\) from \(NaCNO\) ) is \(c_{A^{-}}=\frac{n_{A^{-}}}{V}\), where \(n_{A^{-}} = 0.410\space mol\) and \(V = 2.0\space L\), so \(c_{A^{-}}=\frac{0.410}{2.0}=0.205\space M\).

Step2: Substitute into Henderson - Hasselbalch equation

Substitute \(pK_{a}=3.456\), \([A^{-}]=0.205\space M\) and \([HA]=0.065\space M\) into \(pH = pK_{a}+\log\frac{[A^{-}]}{[HA]}\)

$$
LATEXBLOCK1
$$

Answer:

\(3.96\)