Henderson-Hasselbalch equation.
$\overline{){\mathbf{pH}}{\mathbf{=}}{{\mathbf{pK}}}_{{\mathbf{a}}}{\mathbf{+}}{\mathbf{log}}{\mathbf{}}\mathbf{\left(}\frac{\mathbf{conjugate}\mathbf{}\mathbf{base}}{\mathbf{weak}\mathbf{}\mathbf{acid}}\mathbf{\right)}}$
• HF is a weak acid
• the anion in NaF is the conjugate base
▪ F^{-} → anion → conjugate base
- concentration of F^{-} = concentration of NaF
weak acid + conjugate base = buffer
Step 1: Calculate pK_{a}
K_{a} of HF = 7.24 x 10^{-4} (can be found in books or online)
$\overline{){{\mathbf{pK}}}_{{\mathbf{a}}}{\mathbf{=}}{\mathbf{-}}{\mathbf{log}}{\mathbf{\hspace{0.17em}}}{{\mathbf{K}}}_{{\mathbf{a}}}}\phantom{\rule{0ex}{0ex}}{\mathbf{pK}}_{\mathbf{a}}\mathbf{=}\mathbf{-}\mathbf{log}\mathbf{}\mathbf{(}\mathbf{7}\mathbf{\times}{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{)}$
pK_{a} = 3.140
Step 2: Calculate pH
$\overline{){\mathbf{pH}}{\mathbf{=}}{\mathbf{-}}{\mathbf{log}}{\mathbf{}}{\mathbf{\left[}}{{\mathbf{H}}}_{{\mathbf{3}}}{{\mathbf{O}}}^{{\mathbf{+}}}{\mathbf{\right]}}}\phantom{\rule{0ex}{0ex}}\mathbf{pH}\mathbf{=}\mathbf{-}\mathbf{log}\mathbf{}\mathbf{(}\mathbf{2}\mathbf{.}\mathbf{3}\mathbf{\times}{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{)}$
pH = 3.638
Step 3: Calculate the concentration of NaF.
Conjugate base = [F^{-}] = [NaF]
Weak acid = [HF] = 0.300 M
What concentration of NaF is required to make [H_{3}O^{+}] = 2.3 × 10^{−4} in a 0.300-M solution of HF?
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