Question

Calculate the pH of a solution prepared by mixing $\mathbf{0}\mathbf{.}\mathbf{0800}\mathbf{mol}$ of chloroacetic acid plus$\mathbf{0}\mathbf{.}\mathbf{0400}\mathbf{mol}$ of sodium chloroacetate in$\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{L}$ of water.

(a) First do the calculation by assuming that the concentrations of HA and${\mathbf{A}}^{\mathbf{-}}$ equal their formal concentrations.

(b) Then do the calculation, using the real values of and in the solution.

(c) Using first your head, and then the Henderson-Hasselbalch equation, find the pH of a solution prepared by dissolving all the following compounds in one beaker containing a total volume of$\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{L}\mathbf{:}\mathbf{0}\mathbf{.}\mathbf{180}{\mathbf{molCICHCOO}}_{\mathbf{2}}\mathbf{H}\mathbf{,}\mathbf{0}\mathbf{.}\mathbf{020}\mathbf{}\mathbf{mol}\mathbf{}{\mathbf{CICHCHO}}_{\mathbf{2}}{\mathbf{Na}}_{\mathbf{2}}\mathbf{}\mathbf{0}\mathbf{.}\mathbf{080}{\mathbf{molHNO}}_{\mathbf{3}\mathbf{,}}$and$\mathbf{0}\mathbf{.}\mathbf{080}\mathbf{molCa}{\mathbf{\left(}\mathbf{OH}\mathbf{\right)}}_{\mathbf{2}}$. Assume that$\mathbf{Ca}{\mathbf{\left(}\mathbf{OH}\mathbf{\right)}}_{\mathbf{2}}$ dissociates completely.

Short answer

a) The pH is $2.56$.

b) The pH is$2.61$ .

c) The pH is $2.86$.

Step-by-step solution

Concept used.

The pH of the buffer determined by the concentration ratio of $\left[{\mathrm{A}}^{-}\right]$and $\left[\mathrm{HA}\right]$ by the following Henderson-Hasselbalch equation:$\mathrm{pH}={\mathrm{pK}}_{\mathrm{a}}+\log \left(\left[{\mathrm{A}}^{-}\right]/\left[\mathrm{HA}\right]\right)$

First do the calculation by assuming that the concentrations of HA and A-  equal their formal concentrations.

a)

The pH,

$$\begin{gathered} \mathrm{pH}={\mathrm{pK}}_{\mathrm{a}}+\log \left(\left[{\mathrm{A}}^{-}\right]/\left[\mathrm{HA}\right]\right) \\ \mathrm{pH}={\mathrm{pK}}_{\mathrm{a}}+\log \left(\left[\mathrm{sodium}\mathrm{chloroacetate}\right]/\left[\mathrm{chloroacetic}\mathrm{acod}\right]\right) \\ \mathrm{pH}=2.865+\log \left(0.04/0.08\right) \\ =2.56 \end{gathered}$$

The calculation using the real values of HA and  A- in the solution.

b)

Calculate the pH by using the real values of $\left[\mathrm{HA}\right]$ and $\left[{\mathrm{A}}^{-}\right]$:

$$\begin{gathered} {\mathrm{K}}_{\mathrm{a}}=\frac{\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{A}}^{-}\right]}{\left[\mathrm{HA}\right]} \\ =\frac{\left[{\mathrm{H}}^{+}\right](0.04-\left[{\mathrm{H}}^{+}\right]}{0.08-\left[{\mathrm{H}}^{+}\right]} \\ \left[{\mathrm{H}}^{+}\right]=2.48\times {10}^3\mathrm{M} \\ \mathrm{pH}=-\log \left[{\mathrm{H}}^{+}\right] \\ =-\log \left(2.48\times {10}^3\right) \\ =2.61 \end{gathered}$$

The pH of a solution prepared by dissolving all the following compounds in one beaker.