Question

Calculate the number of moles of$\mathit{H}\mathit{C}\mathit{l}\mathbf{\left(}\mathit{g}\mathbf{\right)}$must be added to$\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{ }\mathit{L}\mathbf{ }\mathit{o}\mathit{f}\mathbf{ }\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{ }\mathit{M}\mathbf{ }\mathit{N}\mathit{a}{\mathit{C}}_{\mathbf{2}}{\mathit{H}}_{\mathbf{3}}{\mathit{O}}_{\mathbf{2}}$to produce a solution buffered at each pH?

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

b)$\mathbf{\text{pH}}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{20}$

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

Short answer

a) To produce a solution buffered at$\text{pH}={\text{pK}}_{\text{a}}$,$\text{0.50} \text{mole} \text{of} \text{HCl}$should be added to$\text{1.0} \text{L} \text{of} \text{the} 1.\text{0M}$ sodium acetate solution.

b) To produce a solution buffered at$\text{pH}=4.20$,$\text{0.78} \text{mole} \text{of} \text{HCl}$should be added to $\text{1.0} \text{L} \text{of} \text{the} 1.\text{0M}$sodium acetate solution.

c) To produce a solution buffered at$\text{pH}=5.00$,$\text{0.36} \text{mole} \text{of} \text{HCl}$should be added to$\text{1.0} \text{L} \text{of} \text{the} 1.\text{0M}$sodium acetate solution.

Step-by-step solution

Definition of mole

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

Calculations

a) $\text{HCl}$reacts with sodium acetate and converts it into acetic acid.

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

$$\begin{gathered} \left[{\text{CH}}_{\text{3}}{\text{COO}}^{\text{-}}\right]=\left[{\text{CH}}_{\text{3}}\text{COOH}\right] \\ =0.50 \text{M} \\ =\left[{\text{H}}^{+}\right]\text{added} \end{gathered}$$

$\text{0.50} \text{mole} \text{of} \text{HCl}$should be added to $\text{1.0} \text{L} \text{of} \text{the} 1.\text{0M}$sodium acetate solution.

b) $\text{HCl}$reacts with sodium acetate and converts it into acetic acid.

$$\begin{gathered} \text{pH}={\text{pK}}_{\text{a}}+\log \frac{\left[{\text{CH}}_{\text{3}}{\text{COO}}^{\text{-}}\right]}{\left[{\text{CH}}_{\text{3}}\text{COOH}\right]} \\ 4.20=4.74+\log \frac{\left[{\text{CH}}_{\text{3}}{\text{COO}}^{\text{-}}\right]}{\left[{\text{CH}}_{\text{3}}\text{COOH}\right]} \end{gathered}$$

$$\begin{gathered} \frac{\left[{\text{CH}}_{\text{3}}{\text{COO}}^{\text{-}}\right]}{\left[{\text{CH}}_{\text{3}}\text{COOH}\right]}={10}^{-0.54}=0.29 \\ \left[{\text{CH}}_{\text{3}}{\text{COO}}^{\text{-}}\right]=0.29\left[{\text{CH}}_{\text{3}}\text{COOH}\right] \\ \left[{\text{CH}}_{\text{3}}{\text{COO}}^{\text{-}}\right]+\left[{\text{CH}}_{\text{3}}\text{COOH}\right]=1.0 \text{M} \end{gathered}$$

$$\begin{gathered} 0.29\left[{\text{CH}}_{\text{3}}\text{COOH}\right]+\left[{\text{CH}}_{\text{3}}\text{COOH}\right]=1.0 \text{M} \\ \left[{\text{CH}}_{\text{3}}\text{COOH}\right]=\frac{1.0}{1.29} \\ =0.78 \text{M}=\left[{\text{H}}^{+}\right]\text{added} \end{gathered}$$

$\text{0.78} \text{mole} \text{of} \text{HCl}$should be added to $\text{1.0} \text{L} \text{of} \text{the} 1.\text{0M}$sodium acetate solution.

c) $\text{HCl}$reacts with sodium acetate and converts it into acetic acid.

$$\begin{gathered} \text{pH}={\text{pK}}_{\text{a}}+\log \frac{\left[{\text{CH}}_{\text{3}}{\text{COO}}^{\text{-}}\right]}{\left[{\text{CH}}_{\text{3}}\text{COOH}\right]} \\ 5.00=4.74+\log \frac{\left[{\text{CH}}_{\text{3}}{\text{COO}}^{\text{-}}\right]}{\left[{\text{CH}}_{\text{3}}\text{COOH}\right]} \end{gathered}$$

$$\begin{gathered} \frac{\left[{\text{CH}}_{\text{3}}{\text{COO}}^{\text{-}}\right]}{\left[{\text{CH}}_{\text{3}}\text{COOH}\right]}={10}^{0.26}=1.8 \\ \left[{\text{CH}}_{\text{3}}{\text{COO}}^{\text{-}}\right]=1.8\left[{\text{CH}}_{\text{3}}\text{COOH}\right] \\ \left[{\text{CH}}_{\text{3}}\text{COOH}\right]=0.56\left[{\text{CH}}_{\text{3}}{\text{COO}}^{\text{-}}\right] \end{gathered}$$

$$\begin{gathered} \left[{\text{CH}}_{\text{3}}{\text{COO}}^{\text{-}}\right]+\left[{\text{CH}}_{\text{3}}\text{COOH}\right]=2.0 \text{M} \\ \left[{\text{CH}}_{\text{3}}{\text{COO}}^{\text{-}}\right]+0.56\left[{\text{CH}}_{\text{3}}{\text{COO}}^{-}\right]=2.0 \text{M} \\ \left[{\text{CH}}_{\text{3}}{\text{COO}}^{\text{-}}\right]=\frac{2.0}{1.56} \\ =1.30 \text{M}=\left[{\text{OH}}^{\text{-}}\right]\text{added} \end{gathered}$$

$\text{0.36} \text{mole} \text{of} \text{HCl}$should be added to$\text{1.0} \text{L} \text{of} \text{the} 1.\text{0M}$sodium acetate solution.