Chapter 18.4, Problem 45PP

Interpretation Introduction

Interpretation:

Molarity of given ammonia solution is to be calculated when it is neutralized by acid.

Concept introduction:

Molarity is number of moles of solute present per liter of solution.

When acid and base are mixed, they neutralize each other. So their molar concentrations can be calculated by applying mole concept.

Answer to Problem 45PP

Molarity of ammonia is $1.178M$

Explanation of Solution

Given data:

$\begin{array}{l}{V}_{HCl}=49.90mL\\ MolarityofHCl=0.59\text{}M\\ V_{N{H}_3}=25mL\end{array}$

$\begin{array}{c}{n}_{HCl}=Molarity\times volumeinliters\\ =0.59\times \frac{49.90}{1000}\\ =2.944\times {10}^{-2}mol\end{array}$

When hydrochloric acid reacts with ammonia solution reaction is given by:

$N{H}_3+HCl\to N{H}_4{}^{+}+C{l}^{-}$

Thus one mol acid reacts with one mol base.

This impies,

$n_{N{H}_3}=2.944\times {10}^{-2}mol$

$\begin{array}{c}MolarityofN{H}_{3}solution=\frac{2.944\times {10}^{-2}mol}{25/1000}\\ =1.178M\end{array}$

Conclusion

$MolarityofN{H}_3$ is $1.178M$