Question

How do you calculate the pH at the equivalence point for the titration of $.190M$ methylamine with $.190MHCl$ ? The Kb of methylamine is $5.0 \times {10^{ - 4}}$ .

Answer 1

Hint: We need to remember that the $pH$ is a measure used in chemistry to specify the acidity or fundamentality of an aqueous solution. To have lower pH values than basic or alkaline solutions, acidic solutions are tested. The pH scale is logarithmic and displays the concentration of hydrogen ions in the solution inversely.
To determine the concentration of an identified analyte, titration is a popular laboratory method of quantitative chemical analysis. As a regular solution of known concentration and volume, a reagent, called the titrant or titrator, is prepared.

Complete step by step answer:
\[pH = 5.86\]
Explanation:
The net ionic equation for the titration in question is the following:
\[C{H_3}N{H_2} + {H^ + } \to C{H_3}NH_3^ + \]
This exercise will be solved suing two kinds of problems: Stoichiometry problem & equilibrium problem .
Stoichiometry Problem :
At the equivalence point, the number of mole of an acid added is equal to the number ofmole of base present.
Since the concentrations of base and acid are equal, concentration of the conjugate acid \[C{H_3}NH_3^ + \] can be determined as follows:
Since equal volumes of the acid and base should be mixed, and since they are additive, the concentration of \[C{H_3}NH_3^ + \]will be half the initial concentration of \[C{H_3}N{H_2}\].
Thus, \[\left[ {C{H_3}NH_3^ + } \right]\]=\[0.095M\]
Equilibrium Problem :
The conjugate acid that will be the major species at the equivalence point, will be the only significant source of H+ in the solution and therefore, to find the pH of the solution we should find the [H+] from the dissociation of \[C{H_3}NH_3^ + \]:
\[C{H_3}NH_3^ + \rightleftarrows C{H_3}N{H_2} + {H^ + }\]
\[
  \begin{array}{*{20}{c}}
  {Initial}&{{\text{ }}0.095M}&{0M}&{0M}
\end{array} \\
  \begin{array}{*{20}{c}}
  {Change}{{\text{ }} - xM}&{{\text{ }} + xM}&{ + xM}
\end{array} \\
  \begin{array}{*{20}{c}}
  {Equilibrium}{\left( {0.095 - x} \right)M}&{xM}&{xM}
\end{array} \\
 \]
\[\;{K_a} = [C{H_3}N{H_2}][{H^ + }][C{H_3}N{H^{ + 3}}]\]\[\]
\[{K_w} = {K_a} \times {K_b}\]
${K_a} = \dfrac{{{K_w}}}{{{K_b}}}$
Now we can substitute the known values and on simplification we get,
$ \Rightarrow {K_a} = \dfrac{{1.0 \times {{10}^{ - 14}}}}{{5.0 \times {{10}^{ - 4}}}} = 2.0 \times {10^{ - 11}}$
$ \Rightarrow {K_a} = \dfrac{{\left[ {C{H_3}N{H_2}} \right]\left[ {{H^ + }} \right]}}{{\left[ {C{H_3}NH_3^ + } \right]}}$
Now we can substitute the known values we get,
$ \Rightarrow {K_a} = \dfrac{{x.x}}{{0.095 - x}} = \dfrac{{{x^2}}}{{0.095 - x}}$
On simplification we get,
${K_a} = 2.0 \times {10^{ - 11}}$
Solve for \[x = 1.38 \times {10^{ - 6}}M = [{H^ + }]\]
Therefore, the pH of the solution is
$pH = - \log \left[ {{H^ + }} \right]$
\[ \Rightarrow pH = - log(1.38 \times {10^{ - 6}}) = 5.86\]

Note:
Using a transfer concentration cell, primary pH standard values are calculated by calculating the potential difference between a hydrogen electrode and a standard electrode, such as the silver chloride electrode. A glass electrode and pH metre, or color-changing indicator, may be used to measure the pH of aqueous solutions.