Question

How many moles of $\mathit{N}\mathit{a}\mathit{O}\mathit{H}$must be added to$\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{ }\mathit{L}\mathbf{ }\mathit{o}\mathit{f}\mathbf{ }\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{ }\mathit{M}\mathbf{ }\mathit{H}{\mathit{C}}_{\mathbf{2}}{\mathit{H}}_{\mathbf{3}}{\mathit{O}}_{\mathbf{2}}$to produce a solution buffered at each pH?

a)$\mathbf{\text{pH}}\mathbf{=}{\mathbf{\text{pK}}}_{\mathbf{\text{a}}}$

b)$\mathbf{\text{pH}}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{00}$

c)$\mathbf{\text{pH}}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{00}$

Short answer

a) To produce a solution buffered at $\text{pH}={\text{pK}}_{\text{a}}$,$\text{2.0} \text{mole} \text{of} \text{NaOH}$should be added to$\text{1.0} \text{L} \text{of} \text{the} \text{2.0M}$acetic acid solution.

b) To produce a solution buffered at$\text{pH}=4.00$ ,$0.3 \text{mole} \text{of} \text{NaOH}$should be added to$\text{1.0} \text{L} \text{of} \text{the} \text{2.0M}$acetic acid solution.

c) To produce a solution buffered at$\text{pH}=4.00$ ,$1.3 \text{mole} \text{of} \text{NaOH}$should be added to$\text{1.0} \text{L} \text{of} \text{the} \text{2.0M}$acetic acid solution.

Step-by-step solution

Definition of mole

An SI unit used to measure the amount of any substance is defined as the mole.

Calculations

a) $\text{NaOH}$reacts with acetic acid and forms sodium acetate and water.

$$\begin{gathered} \text{pH}={\text{pK}}_{\text{a}} \\ \log \frac{\left[{\text{CH}}_{\text{3}}{\text{COO}}^{\text{-}}\right]}{\left[{\text{CH}}_{\text{3}}\text{COOH}\right]}=0 \\ \frac{\left[{\text{CH}}_{\text{3}}{\text{COO}}^{\text{-}}\right]}{\left[{\text{CH}}_{\text{3}}\text{COOH}\right]}=1 \\ \left[{\text{CH}}_{\text{3}}{\text{COO}}^{\text{-}}\right]=\left[{\text{CH}}_{\text{3}}\text{COOH}\right] \\ =2.0 \text{M} \\ =\left[{\text{OH}}^{\text{-}}\right] \end{gathered}$$

$\text{2.0} \text{mole} \text{of} \text{NaOH}$should be added to$\text{1.0} \text{L} \text{of} \text{the} \text{2.0M}$acetic acid solution.

b) $\text{NaOH}$reacts with acetic acid and forms sodium acetate and water.

$$\begin{gathered} \text{pH}={\text{pK}}_{\text{a}}+log\frac{\left[{\text{CH}}_{\text{3}}{\text{COO}}^{\text{-}}\right]}{\left[{\text{CH}}_{\text{3}}\text{COOH}\right]} \\ 4.00=4.74+log\frac{\left[{\text{CH}}_{\text{3}}{\text{COO}}^{\text{-}}\right]}{\left[{\text{CH}}_{\text{3}}\text{COOH}\right]} \end{gathered}$$

$$\begin{gathered} \frac{\left[{\text{CH}}_{\text{3}}{\text{COO}}^{\text{-}}\right]}{\left[{\text{CH}}_{\text{3}}\text{COOH}\right]}={10}^{-0.74}=0.18 \\ \left[{\text{CH}}_{\text{3}}{\text{COO}}^{\text{-}}\right]=0.18\left[{\text{CH}}_{\text{3}}\text{COOH}\right] \\ \left[{\text{CH}}_{\text{3}}\text{COOH}\right]=5.6\left[{\text{CH}}_{\text{3}}{\text{COO}}^{\text{-}}\right] \\ \left[{\text{CH}}_{\text{3}}{\text{COO}}^{\text{-}}\right]+\left[{\text{CH}}_{\text{3}}\text{COOH}\right]=2.0 \text{M} \\ \left[{\text{CH}}_{\text{3}}{\text{COO}}^{\text{-}}\right]+5.6\left[{\text{CH}}_{\text{3}}{\text{COO}}^{-}\right]=2.0 \text{M} \end{gathered}$$

$$\begin{gathered} \left[{\text{CH}}_{\text{3}}{\text{COO}}^{\text{-}}\right]=\frac{2.0}{6.6} \\ =0.30 \text{M}=\left[{\text{OH}}^{\text{-}}\right]\text{added} \end{gathered}$$

$\text{0.3} \text{mole} \text{of} \text{NaOH}$should be added to$\text{1.0} \text{L} \text{of} \text{the} \text{2.0M}$acetic acid solution.

c)$\text{NaOH}$ reacts with acetic acid and forms sodium acetate and water.

$$\begin{gathered} pH=p{K}_a+\log \frac{\left[C{H}_{3}CO{O}^{-}\right]}{\left[C{H}_{3}COOH\right]} \\ 5.00=4.74+\log \frac{\left[C{H}_{3}CO{O}^{-}\right]}{\left[C{H}_{3}COOH\right]} \end{gathered}$$

$$\begin{gathered} \frac{\left[{\text{CH}}_{\text{3}}{\text{COO}}^{\text{-}}\right]}{\left[{\text{CH}}_{\text{3}}\text{COOH}\right]}={10}^{0.26}=1.8 \\ \left[{\text{CH}}_{\text{3}}{\text{COO}}^{\text{-}}\right]=1.8\left[{\text{CH}}_{\text{3}}\text{COOH}\right] \\ \left[{\text{CH}}_{\text{3}}{\text{COO}}^{\text{-}}\right]+\left[{\text{CH}}_{\text{3}}\text{COOH}\right]=1.0\text{M} \end{gathered}$$

$$\begin{gathered} 1.8\left[{\text{CH}}_{\text{3}}\text{COOH}\right]+\left[{\text{CH}}_{\text{3}}\text{COOH}\right]=1.0\text{M} \\ \left[{\text{CH}}_{\text{3}}\text{COOH}\right]=\frac{1.0}{2.8} \\ =0.36 \text{M}=\left[{\text{OH}}^{\text{-}}\right]\text{added} \end{gathered}$$

$1.3 \text{mole} \text{of} \text{NaOH}$should be added to$\text{1.0} \text{L} \text{of} \text{the} \text{2.0M}$acetic acid solution.