Chapter 16, Problem 6STP

Interpretation Introduction

Interpretation: The pH of a 0.045 M solution of nitric acid is to be calculated.

Concept introduction: The $[{\text{H}}^{+}]$ in an aqueous solution can be used to calculate pH,

The pH is defined as

  $\text{pH}=-\text{log}[{\text{H}}^{+}]$

Answer to Problem 6STP

C

Explanation of Solution

Given information,

  $\text{0}\text{.045M}\text{solution}\text{of}{\text{HNO}}_3$

${\text{HNO}}_3$ is a strong acid, so the ions in solutions are ${\text{H}}^{+}$ and ${\text{NO}}_3{}^{-}$ . So,

  $\text{0}\text{.045M}{\text{HNO}}_3\to \text{0}\text{.045M}{\text{H}}^{\text{+}}\text{and}\text{0}\text{.045M}{\text{HNO}}_{\text{3}}{}^{\text{-}}$

Thus, $\begin{array}{l}{\text{[H}}^{\text{+}}\text{]}=\text{0}\text{.045}\text{M}\\ \text{and}\text{pH}=-{\text{log(H}}^{\text{+}}\text{)}\\ \text{=}-\text{log}(0.045)\\ =1.35\end{array}$

Conclusion

Mathematical calculations show that option C is correct.