Chapter U6.7, Problem 4E

(a)

Interpretation Introduction

Interpretation :

The value of K_a of HF must be determined.

Concept Introduction :

Acid dissociation constant (Ka) is small for weak acid. It is the ratio of concentration of products to reactants.

For a weak acid HA, K_a can be written as

  ${\text{K}}_{\text{a}}\text{=}\frac{\text{[H+][A-]}}{\text{[HA]}}$ , where [H+], [A-] and [HA] are concentrations of proton, counter anion and the acid respectively.

Answer to Problem 4E

Ka for HF is $6.94\times {10}^{-5}.$

Explanation of Solution

  $\begin{array}{l}{\text{K}}_{\text{a}}\text{=}\frac{\text{[H+][A-]}}{\text{[HA]}}=\frac{[2.6\times {10}^{-3}][2.6\times {10}^{-3}]}{0.0974}\\ =\frac{6.76\times {10}^{-6}}{0.0974}=6.94\times {10}^{-5}.\end{array}$

b)

Interpretation Introduction

Interpretation :

The value of K_a for HNO₃ is large must be shown.

Concept Introduction :

Strong acid undergoes complete dissociation.

Answer to Problem 4E

The K_a of HNO₃ is infinite.

Explanation of Solution

  ${\text{K}}_{\text{a}}\text{=}\frac{{\text{[H}}^{\text{+}}{\text{][NO}}_{\text{3}}{}^{\text{-}}\text{]}}{{\text{[HNO}}_{\text{3}}\text{]}}=\frac{[0.10][0.10]}{[0]}=\infty$

c)

Interpretation Introduction

Interpretation :

The stronger acid from HF or HNO₃ must be determined.

Concept Introduction :

A strong acid undergoes complete dissociation whereas a weak acid undergo partial dissociation.

Answer to Problem 4E

HNO₃ is stronger acid.

Explanation of Solution

HNO₃ has acid dissociation constant very large which is infinite.

HF is weaker acid as acid dissociation constant is very less which is $6.94\times {10}^{-5}.$

d)

Interpretation Introduction

Interpretation :

The solution with higher concentration of H⁺(aq) 0.10 M HF or 0.10 M HOCl must be explained.

Concept Introduction :

An acid will produce more H+ in aqueous solution which has greater Ka value.

Answer to Problem 4E

0.10 M HF solution has higher concentration of H⁺(aq).

Explanation of Solution

HF has acid dissociation constant $6.94\times {10}^{-5}.$

HOCl has acid dissociation constant $2.8\times {10}^{-8}.$

Thus HF will furnish more H+ in aqueous solution.