Chemistry - Ionic Equilibrium Question with Solution | TestHub

Because concentration of $\mathrm{H}+$ from weak acid is less we need to consider self ionization of ${\mathrm{H}}_2\mathrm{O}$ also. $\begin{array}{rl}\mathrm{HX}(\mathrm{aq})\rightleftharpoons & {\mathrm{H}}^{+}+{\mathrm{X}}^{-}(\mathrm{aq})\\ 10^{-3}-\mathrm{x}& \mathrm{x}+\mathrm{y}\mathrm{x}\\ {\mathrm{H}}_2\mathrm{O}(l)\rightleftharpoons & {\mathrm{H}}^{+}(\mathrm{aq})+{\mathrm{OH}}^{-}(\mathrm{aq})\\ & \mathrm{x}+\mathrm{y}\mathrm{y}\end{array}$ Approximation : $\left(10^{-3}-\mathrm{x}\right)\simeq 10^{-3}$ $\begin{array}{rl}& \Rightarrow \frac{x(x+y)}{10^{-3}}=K_a=4\times 10^{-11}....(1)\\ & \Rightarrow y(x+y)=Kw=10^{-14}.....(2)\end{array}$ $\begin{array}{l}\text{ Add }(1)+(2)\end{array}$ $\begin{array}{ll}\Rightarrow & (\mathrm{x}+\mathrm{y}{)}^2=5\times 10^{-14}\\ \Rightarrow & \mathrm{x}+\mathrm{y}=\left[{\mathrm{H}}^{+}\right]=\sqrt{5}\times 10^{-7}\\ \Rightarrow & \mathrm{x}=\sqrt{5}=2.236\end{array}$ Answer 2.23 or 2.24.