Question

A 100 mL sample of 0.10 M $N{H_3}$ is titrated with 0.10 M $HN{O_3}$. How do you determine the pH of the solution after adding of 40.0 mL of $HN{O_3}$? ${K_b}$ of $N{H_3}$ is $1.8 \times {10^{ - 5}}$?

Answer 1

Hint: In this question, base is reacting with acid so the resulting reaction in neutralization reaction. First we need to find out the pOH of the base with the help of the expression $pOH = \dfrac{1}{2}(p{K_b} - \log b)$ where $p{K_b}$ is the base dissociation constant and b is the concentration of base.

Complete step by step answer:
Given,
The volume of ammonia is 100 mL.
The molarity of ammonia is 0.10 M.
The molarity of nitric acid is 0.10 M.
Volume of nitric acid is 40.0 mL.
${K_b}$ is $1.8 \times {10^{ - 5}}$
Here, the neutralization reaction is taking place as ammonia ($N{H_3}$) is reacted with nitric acid ($HN{O_3}$). Ammonia is the base and nitric acid is the acid.
The reaction involved is shown below.
$N{H_3}(aq) + HN{O_3}(aq) \to N{H_4}N{O_3}(aq) + {H_2}O(l)$
In this reaction, ammonia reacts with nitric acid to form ammonium nitrate and water.
The moles are calculated by the formula shown below.
$n = c \times V$
Where,
n is the number of moles
c is the concentration
V is the volume.
To calculate the moles of ammonia substitute the value in the above equation.
$ \Rightarrow n = 0.10 \times 0.1L$
$ \Rightarrow n = 0.01mol$
To calculate the moles of nitric acid substitute the value in the above equation.
$ \Rightarrow n = 0.10 \times 0.04L$
$ \Rightarrow n = 0.004mol$
The molar ratio by which the acid and base reacts is 1:1.
The remaining number of moles of ammonia is $ \Rightarrow 0.01 - 0.004 = 0.006$.
The total volume will be $ \Rightarrow 100.0 + 40.0 = 140.0$
To calculate the concentration of ammonia, substitute the values in formula.
$ \Rightarrow C = \dfrac{{0.006}}{{0.14}}$
$ \Rightarrow C = 0.04286$mol/L.
From the ICE table, we get an expression.
$pOH = \dfrac{1}{2}(p{K_b} - \log b)$
Where,
b is the concentration of base.
As we know that,
$p{K_b} = - \log {K_b}$
Substitute the value of ${K_b}$
$ \Rightarrow p{K_b} = - \log (1.8 \times {10^{ - 5}})$
$ \Rightarrow p{K_b} = 4.744$
Substitute the value of $p{K_b}$ and concentration of base to find pOH.
$ \Rightarrow pOH = \dfrac{1}{2}(4.744 - \log (0.04286))$
$ \Rightarrow pOH = 3.056$
The relation between pH and pOH is shown below.
$pH + pOH = 14$
Substitute the value of pOH to find the pH.
$ \Rightarrow pH = 14 - 3.056$
$ \Rightarrow pH = 10.9$
Therefore, the pH of the solution is 10.9.

Note:
Make sure to convert the value of volume in milliliter into liters as concentration is given in terms of mol/L. The relation between the pH and pOH can also be written as shown below.
$pH + pOH = {k_w}$
Where ${k_w}$ is the constant used for water.
For water, the pH and pOH value is equal to 7.