Question

0.15 mole of pyridium chloride has been added into $500{\text{ c}}{{\text{m}}^{\text{3}}}$ of 0.2 M pyridine solution. Calculate the ${\text{pH}}$ and hydroxyl ion concentration in the resulting solution assuming no change in volume. ${\text{(}}{{\text{K}}_{\text{b}}}{\text{ for pyridine = 1}}{\text{.5}} \times {\text{1}}{{\text{0}}^{{\text{ - 9}}}}{\text{M)}}$

Answer 1

A solution of pyridine and pyridinium chloride will form a buffer system. Pyridine is a weak base. Write the reaction of pyridine with water. Calculate the concentration of pyridinium chloride added using given moles of it and volume of solution. Use the Henderson Hassel balch equation to calculate the ${\text{pH}}$ of the solution. Using ${\text{pH}}$of solution calculate the ${\text{pOH}}$ of solution. From the${\text{pOH}}$ of the solution calculate the concentration of hydroxide ion.

Complete solution:
As pyridine is a weak base it abstracts protons in an aqueous medium and is converted into pyridinium ion.
${{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{N + }}{{\text{H}}_{\text{2}}}{\text{O }} \to {\text{ }}{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{N}}{{\text{H}}^{\text{ + }}}{\text{ + O}}{{\text{H}}^{\text{ - }}}$
Pyridine pyridinium ion.
${{\text{K}}_{\text{b}}}$ value of pyridine given to us is${\text{1}}{\text{.5}} \times {\text{1}}{{\text{0}}^{{\text{ - 9}}}}$ that indicates that pyridine is a weak base. So the concentration of reacted pyridinium will be negligible. So we can consider the equilibrium concentration of pyridine as the initial concentration of pyridine and the equilibrium concentration of pyridinium ion is nothing but the concentration of pyridinium chloride added.
So, we can say that the equilibrium concentration of pyridine [${{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{N }}$] is 0.2M.
We can calculate the equilibrium concentration of pyridinium ion as follows:
$[{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{N}}{{\text{H}}^{\text{ + }}}] = \dfrac{{{\text{0}}{\text{.15mol}}}}{{{\text{0}}{\text{.5L}}}} = 0.3{\text{M}}$
Now use the Henderson Hasselbalch equation and calculate ${\text{pH}}$ of the solution.
${\text{pH = p}}{{\text{K}}_{\text{a}}}{\text{ + log}}\dfrac{{{\text{[Base]}}}}{{{\text{[Acid]}}}}$
We can rewrite it as
${\text{pH = p}}{{\text{K}}_{\text{a}}}{\text{ + log}}\dfrac{{{\text{[}}{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{N ]}}}}{{{\text{[}}{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{N}}{{\text{H}}^{\text{ + }}}{\text{]}}}}$
Here we need to calculate ${\text{p}}{{\text{K}}_a}$ from the given ${{\text{K}}_{\text{b}}}$ value of ammonium hydroxide.
${\text{p}}{{\text{K}}_{\text{b}}} = - \log ({{\text{K}}_{\text{b}}}) = - \log ({\text{1}}{\text{.5}} \times {\text{1}}{{\text{0}}^{{\text{ - 9}}}}) = 8.82$
Using the relation between ${\text{p}}{{\text{K}}_{\text{b}}}$ and ${\text{p}}{{\text{K}}_a}$ we can calculate the value of ${\text{p}}{{\text{K}}_a}$ as follows:
${\text{p}}{{\text{K}}_a} + {\text{p}}{{\text{K}}_{\text{b}}} = 14$
So, ${\text{p}}{{\text{K}}_a} = 14 - {\text{p}}{{\text{K}}_{\text{b}}} = 14 - 8.82 = 5.18$
Now, substitute 5.18 for ${\text{p}}{{\text{K}}_a}$,0.2M for [${{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{N }}$], 0.3M for$[{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{N}}{{\text{H}}^{\text{ + }}}]$ and calculate the${\text{pH}}$ of the solution.
${\text{pH = 5}}{\text{.18 + log}}\dfrac{{{\text{0}}{\text{.2M}}}}{{{\text{0}}{\text{.3M}}}}$
${\text{pH = 5}}{\text{.0}}$
Hence, ${\text{pH}}$ of the solution is 5.0.
Now, using the relation between ${\text{pH}}$ and ${\text{pOH}}$ of solution calculate the ${\text{pOH}}$ of solution.
${\text{pOH + pH = 14}}$
${\text{pOH = 14 - pH}}$
${\text{pOH = 14 - 5 = 9}}$
Now, calculate the concentration of hydroxide ion using ${\text{pOH}}$ of solution.
${\text{pOH = }} - \log [O{H^ - }]$
${\text{9 = }} - \log [O{H^ - }]$
$[O{H^ - }] = antilog( - 9) = 1.0 \times {10^{ - 9}}M$

Hence, ${\text{pH}}$ of the solution is 5.0 and the concentration of hydroxide ion is $1.0 \times {10^{ - 9}}M$.

Note: A mixture of pyridine and pyridinium chloride forms a basic buffer. A basic buffer is a mixture of a weak base and its salt of a strong acid. Buffer resists the change in pH. This problem can be solved using the ICE table method but the Henderson Hasselbaltch method is the easiest method to solve these types of problems.