91影视

Solve for the concentration of each molecule in the beaker.

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

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

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

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

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

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

Write the reaction expression.

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

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

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

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

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

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

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

Red white is_{${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\text{,1脳0.010尘辞濒=0.010尘辞濒}$}

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

Write the reaction expression.

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

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

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

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