Question

Calculate the $pH$ of an aqueous solution of $1M$ ammonium formate, assuming complete dissociation.
$\left( {p{K_a} = 3.8,p{K_b} = 4.8} \right)$

Answer 1

$pH$ is the measure of concentration of ${H^ + }$ ions in a solution. This concentration is different for different substances. For acids $pH$lies between $0 - 7$ and for bases $pH$ lies between $7 - 14$. Aqueous solution is formed when a substance is dissolved in water. This means in the given question ammonium formate is dissolved in water.
Formula used: $pH = 7 + \dfrac{1}{2}\left( {p{K_a} - p{K_b}} \right)$
Where $p{K_a}$ is ionization constant of acid and $p{K_b}$ is ionization constant of base.

Complete step by step answer:
We know $pH$ is a measure of acidity or basicity of a substance. In this question we have given a molarity of ammonium formate solution and ionization constant of acid and base it is made up of. Chemical formula of ammonium formate is $N{H_4}HC{O_2}$ . This is ammonium salt of formic acid. It is made by bubbling ammonia through formic acid. Chemical formula of ammonia is $N{H_3}$ and formic acid is $HCOOH$ . Chemical name of formic acid is methanoic acid. It is a weak acid. Ammonium formate is salt of weak acid and weak base. $pH$of such salt solutions can be found with the help of formula:
$pH = 7 + \dfrac{1}{2}\left( {p{K_a} - p{K_b}} \right)$
Where $p{K_a}$ is ionization constant of acid and $p{K_b}$ is ionization constant of base.
In this question we have given that $p{K_a} = 3.8$ and $p{K_b} = 4.8$.
Substituting these values in the above formula we get:
$pH = 7 + \dfrac{1}{2}\left( {3.8 - 4.8} \right)$
$pH = 7 + \dfrac{1}{2}\left( { - 1.0} \right)$
Solving this we get:
$pH = 6.5$
So, $pH$ of the given solution of ammonium formate is $6.5$.

Note:
Sometimes ${K_a}$ or ${K_b}$ is given in place of $p{K_a}$ or $p{K_b}$ . The relation between $p{K_a}$ or $p{K_b}$ and ${K_a}$ or ${K_b}$ is $p{K_a} = - \log {K_a}$ or $p{K_b} = - \log {K_b}$. From this relation we can easily calculate the quantities that are required. Also the formula we used above varies for different kinds of salt solutions.