# Problem: The weak acid HQ has a pKa of 4.89.(b) Calculate the [OH−] of 0.65 M HQ.

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###### FREE Expert Solution

${\mathbf{K}}_{\mathbf{a}}\mathbf{=}{\mathbf{10}}^{\mathbf{-}{\mathbf{pK}}_{\mathbf{a}}}\mathbf{=}{\mathbf{10}}^{\mathbf{-}\mathbf{4}\mathbf{.}\mathbf{89}}\mathbf{=}{\mathbf{1}}{\mathbf{.}}{\mathbf{288}}{\mathbf{×}}{{\mathbf{10}}}^{\mathbf{-}\mathbf{5}}$

Construct an ICE Chart:

$\overline{){{\mathbf{K}}}_{{\mathbf{a}}}{\mathbf{=}}\frac{\mathbf{products}}{\mathbf{reactants}}{\mathbf{=}}\frac{\left[{H}_{3}{O}^{+}\right]\left[{Q}^{-}\right]}{\left[\mathrm{HQ}\right]}}$

$\mathbf{1}\mathbf{.}\mathbf{288}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{5}}\mathbf{=}\frac{\left(x\right)\left(x\right)}{\mathbf{0}\mathbf{.}\mathbf{65}\mathbf{-}\mathbf{x}}$

$\mathbf{1}\mathbf{.}\mathbf{288}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{5}}\mathbf{=}\frac{{\mathbf{x}}^{\mathbf{2}}}{\mathbf{0}\mathbf{.}\mathbf{65}\mathbf{-}\overline{)\mathbf{x}}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left[1.288×{10}^{-5}=\frac{{x}^{2}}{\overline{)0.65}}\right]\overline{)\left[0.65\right]}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\sqrt{{\mathbf{x}}^{\mathbf{2}}}\mathbf{=}\sqrt{\left(1.288×{10}^{-5}\right)\left(0.65\right)}$

###### Problem Details

The weak acid HQ has a pKa of 4.89.

(b) Calculate the [OH] of 0.65 M HQ.