Chapter 16, Problem 44A

Interpretation Introduction

Interpretation:

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

Concept introduction:

Molarity is number of moles of solute present per litre 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 44A

Molarity of nitric acid is $0.106M$

Explanation of Solution

Given Information:

  $\begin{array}{l}{V}_{NaOH}=26.5mL\\ MolarityofNaOH=0.20\text{}M\\ V_{HN{O}_3}=50mL\end{array}$

Calculation:

  $\begin{array}{c}{n}_{NaOH}=Molarity\times volumeinliters\\ =0.20\times \frac{26.5}{1000}\\ =5.3\times {10}^{-3}mol\end{array}$

When nitric acid reacts with potassium hydroxide reaction is given by:

  $HN{O}_3+NaOH\to NaN{O}_3+H_{2}O$

Thus one mol acid reacts with one mol base.

This implies,

  $n_{HN{O}_3}=5.3\times {10}^{-3}mol$

  $\begin{array}{c}MolarityofHN{O}_3=\frac{5.3\times {10}^{-3}}{50/1000}\\ =0.106M\end{array}$

Conclusion

  $MolarityofHN{O}_3$ is $0.106M$ .