Q18.178CP - Chemistry: Molecular Nature Of Matter And Change

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 is0.50 L, and the HB solution is 0.25 L, and each particle represents mol, find the Ka of each acid. Which acid, if either, is stronger?

Step-by-Step Solution

Answer

HB is the stronger acid.

Step 1: Find stronger acid

Solve for the concentration of each molecule in the beaker.

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

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

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

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

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

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

Write the reaction expression.

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

Solve for ${\text{K}}_{\text{a}}.$

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

Step 2: Solve the equation

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

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

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

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

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

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

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.