Question

How to calculate the pH of $0.180{\text{ g}}$ of potassium biphthalate $({\text{p}}{K_a} = 5.4)$ in $50{\text{ mL}}$ of water?

Answer 1

Potassium biphthalate or potassium hydrogen phthalate is an acidic salt. This is because hydrogen phthalate anion will act as a weak acid in aqueous solution. Using the formula of ${\text{pH}}$ for a salt of strong acid and weak base, we can calculate the same.

Formula used:
${\text{pH}} = 7 - \dfrac{1}{2}{\text{p}}{K_b} - \dfrac{1}{2}\log C$

Complete step by step answer:
First, potassium hydrogen phthalate will dissociate into potassium cation and hydrogen phthalate anion. Then, hydrogen phthalate anion will dissociate into hydrogen cation and phthalate anion. Therefore, the nature of aqueous solution formed by dissolving potassium biphthalate in water will be acidic.
To calculate the ${\text{pH}}$of this solution, we will use the formula for salt of strong acid and weak base as:
${\text{pH}} = 7 - \dfrac{1}{2}{\text{p}}{K_b} - \dfrac{1}{2}\log C$, where $C$ is the molarity of solution.
First we will calculate the molar concentration of given salt, potassium biphthalate.
Given mass of salt, $m = 0.180{\text{ g}}$
Chemical formula of salt, potassium biphthalate is ${\text{HOOC}} - {{\text{C}}_6}{{\text{H}}_4} - {\text{COOK}}$or ${{\text{C}}_8}{{\text{H}}_5}{{\text{O}}_4}{\text{K}}$.
We know that, molar mass of ${\text{C}} = 12$, ${\text{H}} = 1$, ${\text{O}} = 16$ and ${\text{K}} = 39$.
Therefore, molar mass of salt, $M = (8 \times 12) + (5 \times 1) + (4 \times 16) + 39$
$= 96 + 5 + 64 + 39$
$= 204{\text{ g mo}}{{\text{l}}^{ - 1}}$
Moles of salt, $n = \dfrac{m}{M}$
Substituting the values of $m$ and $M$ from above, we get:
$n = \dfrac{{0.180{\text{ g}}}}{{204{\text{ g mo}}{{\text{l}}^{ - 1}}}}$
$\Rightarrow n = \dfrac{{0.18}}{{204}}{\text{ mol}}$
Volume of solution, $V = 50{\text{ mL}}$
$\Rightarrow V = 50 \times {10^{ - 3}}{\text{ L}}$
The molarity of formed solution, $C = \dfrac{n}{V}$
Substituting the values of $n$ and $V$ from above, we get:
$C = \dfrac{\dfrac{0.18}{204}}{50\times10^{-3}}$ $molL^{-1}$
$\Rightarrow C = \dfrac{{0.18}}{{204}} \times \dfrac{{{{10}^3}}}{{50}}{\text{ mol }}{{\text{L}}^{ - 1}}$
$\Rightarrow C = \dfrac{{18}}{{204 \times 5}}{\text{ mol }}{{\text{L}}^{ - 1}}$
Upon solving the above expression:
$C = 0.018{\text{ mol }}{{\text{L}}^{ - 1}}$
It is given that, ${\text{p}}{K_a}$ value of salt is $5.4$.
And we know that, ${\text{p}}{K_a} + {\text{p}}{K_b} = 14$
$\Rightarrow {\text{p}}{K_b} = 14 - 5.4$
$\Rightarrow {\text{p}}{K_b} = 8.6$
Substituting the value of ${\text{p}}{K_b}$ and $C$ in the below expression, we get:
${\text{pH}} = 7 - \dfrac{1}{2}{\text{p}}{K_b} - \dfrac{1}{2}\log C$
$\Rightarrow {\text{pH}} = 7 - \left( {\dfrac{1}{2} \times 8.6} \right) - \left( {\dfrac{1}{2}\log 0.018} \right)$
\Rightarrow {\text{pH}} = 7 - \left( {4.3} \right) - \left\\{ {\dfrac{1}{2}\log \left( {1.8 \times {{10}^{ - 2}}} \right)} \right\\}
$\Rightarrow {\text{pH}} = 7 - \left( {4.3} \right) - \dfrac{1}{2}\left( {\log 1.8 + \log {{10}^{ - 2}}} \right)$
Putting the values of $\log 1.8 = 0.255$ and $\log 10 = 1$:
$\Rightarrow {\text{pH}} = 2.7 + \dfrac{1}{2}\left( {0.255 - 2} \right)$
$\Rightarrow {\text{pH}} = 2.7 + 0.87$
$\Rightarrow {\text{pH}} = 3.57$

Note: The given salt can be confused as a salt of a weak acid phthalic acid and a strong base potassium hydroxide. This will give a wrong value of ${\text{pH}}$. Here, we have to be careful as the hydrogen phthalate ion will produce hydronium ion, which will make the solution weakly acidic.