Question

The value of \[p{K_a}\] of iodic acid is \[\log {\text{ 6}}\] , then find the value of pH of \[1{\text{ M HI}}{{\text{O}}_3}\]solution will be:
A. \[\log {\text{ 6}}\]
B. \[\log {\text{ 5}}\]
C. \[\log {\text{ 4}}\]
D. \[\log {\text{ 3}}\]

Answer 1

Hint: The \[p{K_a}\] of iodic acid is \[\log {\text{ 6}}\]. We will find the value of \[{K_a}\] with the help of \[p{K_a}\] of iodic acid. Then we will find the dissociation of iodic acid when it is dissolved in water to give acidic ions. With the help of concentration hydrogen ions we will find the value of pH of \[1{\text{ M HI}}{{\text{O}}_3}\] solution.

Complete answer:
Since the value of \[p{K_a}\] is given for iodic acid we will find the value of \[{K_a}\] as,
\[p{K_a}{\text{ = - log }}{{\text{K}}_a}\]
\[{\text{6 = - log }}{{\text{K}}_a}\]
Since we cannot find antilog of negative numbers we must convert it into positive value and then taking antilog both sides we get the value of \[{K_a}\]
\[{\text{6 = log }}\dfrac{1}{{{K_a}}}\]
Taking antilog both sides we get,
\[{K_a}{\text{ = 0}}{\text{.167}}\]
Now we will write the acidic reaction of iodic acid when it is dissolved in water.
\[{\text{HI}}{{\text{O}}_3}{\text{ + }}{{\text{H}}_2}{\text{O }} \rightleftharpoons {\text{ }}{{\text{H}}_3}{{\text{O}}^ + }{\text{ + I}}{{\text{O}}^{3 - }}\]
Since the molarity of iodine solution is given as \[1{\text{ M HI}}{{\text{O}}_3}\] we can find the concentration of its ions at \[t{\text{ = 0}}\] and at \[t{\text{ = }}{t_1}\], when dissociation of amount \[x\] starts,
  \[{\text{HI}}{{\text{O}}_3}{\text{ + }}{{\text{H}}_2}{\text{O }} \rightleftharpoons {\text{ }}{{\text{H}}_3}{{\text{O}}^ + }{\text{ + I}}{{\text{O}}^{3 - }}\]

Time(t)	\[\left[ {HI{O_3}} \right]\]	\[\left[ {{H_3}{O^ + }} \right]\]	\[\left[ {I{O_3}^{ - 1}} \right]\]
\[t{\text{ = 0}}\]	1	0	0
\[t{\text{ = }}{t_1}\]	\[1 - x\]	\[x\]	\[x\]

The value of \[{K_a}\] from above data can be written as,
\[{K_a}{\text{ = }}\dfrac{{x{\text{ }} \times {\text{ }}x}}{{1 - x}}\]
\[{K_a}{\text{ = }}\dfrac{{{x^2}}}{{1 - x}}\]
On substituting the value of \[{K_a}\] and ignoring \[1 - x\] as \[x \ll 1\] we can write as,
\[x{\text{ = }}\sqrt {{K_a}} \]
\[x{\text{ = }}\sqrt {0.167} \]
\[x{\text{ = 0}}{\text{.333}}\]
Therefore the concentration of \[{H_3}{O^ + }\] is equal to \[0.333\] and we can find the value of \[pH\] as,
\[pH{\text{ = - log}}\left[ {{H_3}{O^ + }} \right]\]
On substituting the values we get the result as,
\[pH{\text{ = - log}}\left[ {0.333} \right]\]
\[pH{\text{ = log}}\left[ {\dfrac{1}{{0.333}}} \right]\]
\[pH{\text{ = log 3}}\]
Therefore the \[pH\] of \[1{\text{ M HI}}{{\text{O}}_3}\]solution is \[\log {\text{ 3}}\] , thus the correct option is D.

Note:
The base value of all the logarithmic values is ten. We can also use a log table for finding values of logarithmic and antilogarithmic values of various functions. For finding the acidic nature of any acid we make its dissociation in water solvent because the ions get dissociated in water solvent. Also water solvent shows a levelling effect for acids and bases.