Question

The beakers below depict the aqueous dissociations of weak acids HA (blue and green) and HB (blue and yellow); solvent molecules are omitted for clarity. If the HA solution is0.50L, and the HB solution is 0.25L, and each particle representsmol, find the Ka of each acid. Which acid, if either, is stronger?

Short answer

HB is the stronger acid.

Step-by-step solution

Find stronger acid

Solve for the concentration of each molecule in the beaker.

Green is${\text{A}}^{\text{-}}\text{, 1×0.010mol=0.010mol}$

$$\begin{gathered} \left[{\text{A}}^{\text{-}}\right]\text{=}\frac{\text{mol}}{\text{L}} \\ \text{=}\frac{0.010mol}{0.5L} \\ \text{=0.020M.} \end{gathered}$$

Green-white is$\text{HA, 5×0.010mol=0.050mol}$

$$\begin{gathered} \text{[HA] =}\frac{\text{mol}}{\text{L}} \\ \text{=}\frac{0.050mol}{0.5L} \\ =0.10M \end{gathered}$$

Red-white is${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\text{, 1×0.010mol=0.010mol}$

$$\begin{gathered} \left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\text{=}\frac{\text{mol}}{\text{L}} \\ \text{=}\frac{1mol}{0.5L} \\ \text{=0.020M.} \end{gathered}$$

Write the reaction expression.

${\text{HA +H}}_{\text{2}}\text{O}\rightleftharpoons {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{+A}}^{\text{-}}$

Solve for${\text{K}}_{\text{a}}.$

$$\begin{gathered} {\text{K}}_{\text{a}}\text{=}\frac{\text{[ product ]}}{\text{[}reactant\text{]}} \\ \text{=}\frac{\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\left[{\text{A}}^{\text{-}}\right]}{\text{HA]}} \\ \text{=}\frac{\text{(0.020M)(0.020M)}}{\text{0.10M}} \end{gathered}$$

Solve the equation

Yellow is${\text{B}}^{\text{-}}\text{, 1×0.010mol=0.010mol}$

$$\begin{gathered} \left[{\text{B}}^{\text{-}}\right]\text{=}\frac{\text{mol}}{\text{L}} \\ \text{=}\frac{0.010mol}{0.25L} \\ \text{=0.040M} \end{gathered}$$

Green white is$\text{HB, 4×0.010mol=0.040mol}$

$$\begin{gathered} \text{[HB]}\text{=}\frac{\text{mol}}{\text{L}} \\ \text{=}\frac{{\displaystyle 0.040mol}}{0.25L} \\ =0.16\hspace{0.17em}M. \end{gathered}$$

Red white is_{${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\text{,1×0.010mol=0.010mol}$}

$$\begin{gathered} \left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\text{=}\frac{\text{mol}}{\text{L}} \\ \text{=}\frac{1mol}{0.25L} \\ \text{=0.040M.} \end{gathered}$$

Write the reaction expression.

${\text{HB +H}}_{\text{2}}\text{O}\rightleftharpoons {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{+B}}^{\text{-}}$

Solve for${\text{K}}_{\text{a}}$.

$$\begin{gathered} {\text{K}}_{\text{a}}\text{=}\frac{\text{[ product ]}}{\text{[ reactant ]}} \\ \text{=}\frac{\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\left[{\text{B}}^{\text{-}}\right]}{\text{[HB]}} \\ \text{=}\frac{\text{(0.040M)(0.040M)}}{\text{0.16M}} \\ =1.0\times {10}^{-2} \end{gathered}$$

Since the${\text{K}}_{\text{a}}\text{of HB}$is greater than the${\text{K}}_{\text{a}}$of HA, HB is the stronger acid.