**Calculate K**_{b}

$\overline{){{\mathbf{K}}}_{{\mathbf{a}}}{{\mathbf{K}}}_{{\mathbf{b}}}{\mathbf{=}}{{\mathbf{K}}}_{{\mathbf{w}}}}\phantom{\rule{0ex}{0ex}}{\mathbf{K}}_{\mathbf{b}}\mathbf{=}\frac{{\mathbf{K}}_{\mathbf{w}}}{{\mathbf{K}}_{\mathbf{a}}}\phantom{\rule{0ex}{0ex}}{\mathbf{K}}_{\mathbf{b}}\mathbf{=}\frac{\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{\times}{\mathbf{10}}^{\mathbf{-}\mathbf{14}}}{\mathbf{6}\mathbf{.}\mathbf{8}\mathbf{\times}{\mathbf{10}}^{\mathbf{-}\mathbf{4}}}$

**K _{b} = 1.47x10**

Solve an equilibrium problem (using an ICE table) to calculate the pH of each of the following solutions. (K_{a}(HF)=6.8 x 10^{-4})

0.14 M NaF

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