Chapter 17.2, Problem 17.2.1SR

Interpretation Introduction

Interpretation:

The $\text{pH}$ of given solution ( $\text{NaF+}\text{HF}$) has to be calculated.

Concept introduction:

• $\text{pH}$ is the logarithm of the reciprocal of the concentration  of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ in a solution.
• pH of a solution is a measure of concentration of hydrogen ion in a solution. Lower $\text{pH}$ due to more hydrogen ions and higher $\text{pH}$ due to less concentration of hydrogen ions.

To calculate: The $\text{pH}$ of given solution ( $\text{NaF+}\text{HF}$).

Answer to Problem 17.2.1SR

$\text{Option (a) pH}\text{=}3.26$

Explanation of Solution

Reason for correct option(a)

The $\text{pH}$ of given solution ( $\text{NaF+}\text{HF}$) is calculated below.

Given,

$\begin{array}{l}\text{HF}\text{0}\text{.76 M}\\ \\ \text{NaF}\text{0}\text{.98M}\end{array}$

The equilibrium table to find hydrogen ion concentration is

$\begin{array}{l}{\text{HF}}_{\left(\text{aq}\right)}\text{+}{\text{H}}_{\text{2}}\text{O }\stackrel{\rightleftarrows }{\to }{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\left(\text{aq}\right)}\text{+ }{\text{F}}^{\text{-}}{}_{\left(\text{aq}\right)}\\ \text{Initial concentration }\left(\text{M}\right)\text{: 0}\text{.76 }\text{0}\text{0}\text{.98}\\ \text{Change in concentration }\left(\text{M}\right)\text{: }\text{-x }\text{+x }\text{+x}\\ \text{Equilibrium concentration}\left(\text{M}\right)\text{: 0}\text{.76-x}\text{x}\text{0}\text{.98+x}\\ \\ \text{The}{\text{K}}_{\text{a}}\text{of}\text{HF}\text{is}7.1\text{×}{\text{10}}^{\text{-4}}\\ \\ {\text{K}}_{\text{a}}\text{ = }\frac{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{][F}}^{\text{-}}\text{]}}{\text{[HF]}}\\ 7.1\text{×}{\text{10}}^{\text{-4}}\text{= }\frac{\text{(x)(0}\text{.98+x)}}{\text{(0}\text{.76-x)}}\\ \text{x}\text{is}\text{very}\text{small}\text{and}\text{neglect}\text{it,}\text{because}\text{the}\text{ionzation}\text{of}\text{HF}\text{is}\\ \text{suppresssed}\text{by}\text{fluoride}\text{ions}\\ 7.1\text{×}{\text{10}}^{\text{-4}}\text{= }\frac{\text{(x)(0}\text{.98)}}{\text{(0}\text{.76)}}\\ \text{x}\text{=}{\text{[H}}^{\text{+}}\text{]}\text{=}5.50\text{×}{\text{10}}^{\text{-4}}\\ \\ \text{pH = -log (}5.50\text{×}{\text{10}}^{\text{-4}}\text{)}\text{=}3.26\end{array}$

Hence, the correct option is (a)

Reason for incorrect options (b), (c), (d), (e)

The options except (a) have $\text{pH}$ values which is not equal to calculated $\text{pH}$ value of $\text{NaF+}\text{HF}$ buffer solution

Hence, the incorrect options are (b), (c), (d), (e)

Conclusion

The $\text{pH}$ of given solution ( $\text{NaF+}\text{HF}$) was calculated.=