(1) Recall the Henderson-Hasselbalch equation:

$\overline{){\mathbf{pH}}{\mathbf{=}}{{\mathbf{pK}}}_{{\mathbf{a}}}{\mathbf{+}}{\mathbf{log}}\left(\frac{\mathbf{conjugate}\mathbf{}\mathbf{base}}{\mathbf{weak}\mathbf{}\mathbf{acid}}\right)}\phantom{\rule{0ex}{0ex}}\mathbf{pH}\mathbf{=}{\mathbf{pK}}_{{\mathbf{a}}}\mathbf{+}\mathbf{log}{\frac{\left[{\mathrm{KNO}}_{2}\right]}{\left[{\mathrm{HNO}}_{2}\right]}}\phantom{\rule{0ex}{0ex}}\mathbf{pH}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{33}\mathbf{+}\mathbf{log}{\frac{0.15}{0.10}}$

pH = 3.5

(2) Calculating for the moles of NaOH, HNO_{2} and KNO_{2}

$\mathbf{mol}\mathbf{}\mathbf{NaOH}\mathbf{\hspace{0.17em}}\mathbf{=}\mathbf{}\frac{\mathbf{0}\mathbf{.}\mathbf{25}\mathbf{}\mathbf{g}}{\mathbf{39}\mathbf{.}\mathbf{996}\mathbf{}\mathbf{g}\mathbf{/}\mathbf{mol}}$

mol NaOH = 0.0063 moles

mol HNO_{2} = (0.1 M) (0.5 L) = 0.050 moles

mol KNO_{2} = (0.15 M) (0.5 L) = 0.075 moles

A 500.0-mL buffer solution is 0.100 M in HNO_{2} and 0.150 M in KNO_{2}. Determine whether or not 250 mg NaOH would exceed the capacity of the buffer to neutralize it.

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