Problem: A mixture of acids is made that has 0.100 M HCHO 2 and 0.050 M HClO. The Ka for HCHO2 is 1.8 x 10 -4, and the Ka for HClO is 2.9 x 10 -8. Solve for the pH of the mixture of acids.A) 2.37B) 0.82C) 1.00D) 4.42

We are asked to solve for the pH of the mixture of acids.

Calculate the [H₃O⁺] 0.100 M HCHO₂

$$\begin{gathered} \boxed{{\mathbf{K}}_{\mathbf{a}}\mathbf{ }\mathbf{=}\frac{\mathbf{ }\mathbf{products}}{\mathbf{reactants}}} \\ {\mathbf{K}}_{\mathbf{a}}\mathbf{ }\mathbf{=}\frac{\mathbf{\left[}{{\mathbf{CHO}}_{\mathbf{2}}}^{\mathbf{-}}\mathbf{\right]}\mathbf{\left[}{\mathbf{H}}_{\mathbf{3}}{\mathbf{O}}^{\mathbf{+}}\mathbf{\right]}}{\left[{\mathrm{HCHO}}_2\right]} \\ \mathbf{1}\mathbf{.}\mathbf{8}\mathbf{ }\mathbf{\times }\mathbf{ }\mathbf{10}{\mathbf{ }}^{\mathbf{-}\mathbf{4}}\mathbf{ }\mathbf{=}\frac{\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{\left(}\mathbf{x}\mathbf{\right)}}{0.100-x} \end{gathered}$$

Check if x is negligible in the denominator: 

$\frac{0.100}{1.80\times {10}^{-4}} = 555.56 >\hspace{0.17em}500 assume x is negligible$

$$\begin{gathered} \mathbf{1}\mathbf{.}\mathbf{8}\mathbf{ }\mathbf{\times }\mathbf{ }\mathbf{10}{\mathbf{ }}^{\mathbf{-}\mathbf{4}}\mathbf{ }\mathbf{=}\frac{\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{\left(}\mathbf{x}\mathbf{\right)}}{0.100} \\ \mathbf{1}\mathbf{.}\mathbf{8}\mathbf{ }\mathbf{\times }\mathbf{ }\mathbf{10}{\mathbf{ }}^{\mathbf{-}\mathbf{4}}\mathbf{ }\mathbf{=}\frac{{\mathbf{x}}^{\mathbf{2}}}{0.100} \\ {\mathbf{x}}^{\mathbf{2}}\mathbf{ }\mathbf{=}\mathbf{ }(1.8 \times 10{ }^{-4})(0.100) \\ \mathbf{x}\mathbf{ }\mathbf{=}\mathbf{ }\sqrt{\mathbf{(}\mathbf{1}\mathbf{.}\mathbf{8}\mathbf{ }\mathbf{\times }\mathbf{ }\mathbf{10}{\mathbf{ }}^{\mathbf{-}\mathbf{4}}\mathbf{)}(0.100)} \end{gathered}$$

[H₃O⁺] = x = 4.24 x 10^-3 M

Next, do the same procedure for 0.050 M HClO 

$$\begin{gathered} \boxed{{\mathbf{K}}_{\mathbf{a}}\mathbf{ }\mathbf{=}\frac{\mathbf{ }\mathbf{products}}{\mathbf{reactants}}} \\ {\mathbf{K}}_{\mathbf{a}}\mathbf{ }\mathbf{=}\frac{\mathbf{\left[}\mathit{C}\mathit{l}{\mathit{O}}^{\mathbf{-}}\mathbf{\right]}\mathbf{\left[}{\mathbf{H}}_{\mathbf{3}}{\mathbf{O}}^{\mathbf{+}}\mathbf{\right]}}{\left[\mathrm{HClO}\right]} \\ \mathbf{2}\mathbf{.}\mathbf{9}\mathbf{ }\mathbf{\times }\mathbf{ }\mathbf{10}{\mathbf{ }}^{\mathbf{-}\mathbf{8}}\mathbf{ }\mathbf{=}\frac{\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{\left(}\mathbf{x}\mathbf{\right)}}{0.050-x} \end{gathered}$$

Check if x is negligible in the denominator: 

$\frac{0.050}{2.90\times {10}^{-8}} = 1724137.93 >\hspace{0.17em}500 assume x is negligible$

$$\begin{gathered} \mathbf{2}\mathbf{.}\mathbf{9}\mathbf{ }\mathbf{\times }\mathbf{ }\mathbf{10}{\mathbf{ }}^{\mathbf{-}\mathbf{8}}\mathbf{ }\mathbf{=}\frac{\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{\left(}\mathbf{x}\mathbf{\right)}}{0.050} \\ \mathbf{2}\mathbf{.}\mathbf{9}\mathbf{ }\mathbf{\times }\mathbf{ }\mathbf{10}{\mathbf{ }}^{\mathbf{-}\mathbf{8}}\mathbf{ }\mathbf{=}\frac{{\mathbf{x}}^{\mathbf{2}}}{0.050} \\ \mathbf{x}\mathbf{ }\mathbf{=}\mathbf{ }\sqrt{\mathbf{(}\mathbf{2}\mathbf{.}\mathbf{9}\mathbf{ }\mathbf{\times }\mathbf{ }\mathbf{10}{\mathbf{ }}^{\mathbf{-}\mathbf{8}}\mathbf{)}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{050}\mathbf{)}} \end{gathered}$$

[H₃O⁺] = x = 3.8 x 10^-5 M