Question

Question: A 0.050 M solution of the salt NaB has a pH of 9.00. Calculate the pH of a 0.010 M solution of HB.

Short answer

Answer

pH of a 0.010 M solution of HB is 3.6.

Step-by-step solution

Dissociation of NaB

The pH of the solution is given 9 of the 0.050 M solution of NaB, NaB dissociates as:

$$\begin{gathered} {\text{B}}^{-}{\text{+H}}_{\text{2}}\text{O}\rightleftarrows {\text{HB + OH}}^{-} \\ {\text{pH = - log[H}}^{\text{+}}\text{]} \\ {\text{[H}}^{\text{+}}{\text{]=10}}^{\text{- pH}}\backslash \\ \left[O{H}^{-}\right]=\frac{{\text{K}}_{\text{w}}}{{\text{[H}}^{\text{+}}\text{]}} \\ =\frac{1\times {10}^{-14}}{{10}^{-9}} \\ [O{H}^{}-]=1\times {10}^{-5} \end{gathered}$$

[OH^-] = [HB] = 1×10^-5

Ka Calculation

NaB behaves as a base:

$$\begin{gathered} {\text{B}}^{-}{\text{+H}}_{\text{2}}\text{O}\rightleftarrows {\text{HB + OH}}^{-} \\ {\text{K}}_{\text{b}}=\frac{{\text{[HB][OH}}^{\text{-}}\text{]}}{{\text{[B}}^{\text{-}}\text{]}} \\ =\frac{{(1\times {10}^{-5})}^2}{0.07-(1\times {10}^{-5})} \\ {\text{K}}_{\text{b}}=1.4\times {10}^{-9} \\ {\text{K}}_a=\frac{{\text{K}}_w}{{\text{K}}_b}=\frac{1\times {10}^{-14}}{1.4\times {10}^{-9}}=7.14\times {10}^{-6} \end{gathered}$$

pH Calculation for 0.010 M HA acid

K_a for the given acid,

$$\begin{gathered} \text{HB}\rightleftarrows {\text{H}}^{\text{+}}\text{+}{\text{B}}^{-} \\ 0.0100 \\ 0.01-\text{x}\text{x}\text{x}\backslash \\ {\text{K}}_{\text{a}}=\frac{{\text{[H}}^{\text{+}}{\text{][B}}^{\text{-}}\text{]}}{\text{[HB]}}=\frac{\text{[x][x]}}{\text{[0.01 - x]}}=7.14\times {10}^{-6} \\ \text{x}={\text{[H}}^{\text{+}}\text{]}=2.67\times {10}^{-4} \\ \text{pH}=-\log {\text{[H}}^{\text{+}}\text{]}=-\log (2.67\times {10}^{-4})=3.6 \end{gathered}$$