**Step 1: Calculate pOH. **

Use the **Henderson-Hasselbalch** equation:

$\overline{){\mathbf{pOH}}{\mathbf{=}}{{\mathbf{pK}}}_{{\mathbf{b}}}{\mathbf{}}{\mathbf{+}}{\mathbf{}}{\mathbf{log}}\mathbf{\left[}\frac{\mathbf{conjugate}\mathbf{}\mathbf{acid}}{\mathbf{weak}\mathbf{}\mathbf{base}}\mathbf{\right]}}$

$\mathbf{pOH}\mathbf{=}(-\mathrm{log}7.5\times {10}^{-5})\mathbf{}\mathbf{+}\mathbf{}\mathbf{log}\mathbf{\left[}\frac{\mathbf{0}\mathbf{.}\mathbf{530}\mathbf{}\mathbf{M}}{\mathbf{0}\mathbf{.}\mathbf{150}\mathbf{}\mathbf{M}}\mathbf{\right]}\phantom{\rule{0ex}{0ex}}\mathbf{pOH}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{12}\mathbf{}\mathbf{+}\mathbf{}\mathbf{0}\mathbf{.}\mathbf{548}\phantom{\rule{0ex}{0ex}}$

**pOH = 4.67**

If a buffer solution is 0.150 M in a weak base (Kb=7.5x10^{-5}m) and 0.530 M in its conjugate acid, what is the pH?

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