Chapter 14, Problem 14.1P

Interpretation Introduction

To determine:

Rate of consumption of H +

Concept introduction:

Rate of reaction is the decomposition of reactant in particular period of time can be shown as:

For a reaction:

  $\text{A }+\text{ B} \to \text{C}$

Rate can be shown as:

Rate of reaction = $-\text{ d}\left[\text{A}\right]/\text{ dt} =+\text{ d }\left[\text{C}\right]/\text{ dt}$

Where $-\text{ d}\left[\text{A}\right]/\text{ dt}$is rate of reaction with respect to A

Answer to Problem 14.1P

Solution: The consumption of hydrogen ion is = $3.2\times {10}^{-4}\text{M/s}$

Explanation of Solution

The oxidation of Iodine ion by Arsenic acid (H₃AsO₄) follows the below balanced equation -

  ${\text{3I}}^{-}\left(\text{aq}\right) +{\text{ H}}_{\text{3}}{\text{AsO}}_{\text{4}}\left(\text{aq}\right)+{\text{ 2H}}^{+}\left(\text{aq}\right)\to {\text{I}}_{\text{3}}{}^{-}\left(\text{aq}\right)+{\text{ H}}_{\text{3}}{\text{AsO}}_3\left(\text{aq}\right)+{\text{ H}}_{\text{2}}\text{O }\left(\text{l}\right)$

Again, we can correlate the rate equation in terms of change in the concentration of H+ ion.

Rate= $-\frac{1}{3}\frac{\Delta [I-]}{\Delta t}$= $+\frac{\Delta [I{3}^{-}]}{\Delta t}$= $-\frac{1}{2}\frac{\Delta [H^{+}]}{\Delta t}$

So, the consumption of H+ion will be-

$-\frac{\Delta [H^{+}]}{\Delta t}$= $-\frac{2}{3}\frac{\Delta [I^{-}]}{\Delta t}=\frac{2}{3}\times 4.8\times {10}^{-4}\text{M/s}$

= $3.2\times {10}^{-4}\text{M/s}$

Conclusion

Hence,

Rate of production is $=1.6\times {10}^{-4}\text{M/s}$

The consumption of hydrogen ion is = $3.2\times {10}^{-4}\text{M/s}$