Question

The concentration of aqueous ammonia in a cleaning solution is determined by titration with hydrochloric acid. A volume of 23.18 mL of 0.8381 M HCl is needed to titrate a 50.0-mL sample of the ammonia solution to a methyl red endpoint. Calculate the concentration of ammonia in the cleaning solution.

Short answer

The concentration of ammonia in the cleaning solution is $0.389{\text{molL}}^{-1}$.

Step-by-step solution

Given information

Volume of HCl, ${\text{V}}_{\text{HCl}}=23.18\text{mL =}23.18\times {10}^{-3}\text{L}$

Molarity of HCl, ${\text{M}}_{\text{HCl}}=0.8381\text{M}$

Volume of cleaning solution,${\text{V}}_{\text{a}}=50.0\text{mL =}50.0\times {10}^{-3}\text{L}$

Molarity of cleaning solution,${\text{M}}_{\text{a}}=?$

Step 2: Titration

Titration is a process in which a quantified substance is reacted with a non-quantified substance until a chemical equivalence is reached. It helps to calculate several stoichiometric quantities like normality, molarity, molality etc.

Calculation of the concentration of ammonia in cleaning solution

Let us assume that the cleaning solution contains no other bases than ammonia. When equal moles of ammonia and hydrochloric acid reacts with each other, the number of moles of hydrochloric acid to bring the cleaning solution to the end point is same as the number of moles of ammonia present. It can be calculated as follows:

${\text{V}}_{\text{HCl}}\times {\text{M}}_{\text{HCl}}={\text{V}}_{\text{a}}\times {\text{M}}_{\text{a}}$

Substituting the values,

$\begin{array}{rcl}23.18\times {10}^{-3}\text{L}\times 0.8381\text{M}& =& 50.0\times {10}^{-3}\text{L}\times {\text{M}}_{\text{a}}\\ {\text{M}}_{\text{a}}& =& \frac{23.18\times {10}^{-3}\text{L}\times 0.8381\text{M}}{50.0\times {10}^{-3}\text{L}}\\ & =& 0.389\text{M =}0.389{\text{molL}}^{-1}\end{array}$

Hence, the concentration of ammonia in the cleaning solution is role="math" localid="1663351580072" $\mathbf{0}\mathbf{.}\mathbf{389}\mathbf{}{\mathbf{\text{molL}}}^{\mathbf{-}\mathbf{1}}$