Question

At what ${{pH}}$ will a ${{1}}{{{0}}^{{{ - 4}}}}$ M solution of indicator with ${{{K}}_{{b}}}$ = ${{1 \times 1}}{{{0}}^{{{ - 11}}}}$ change colour ?
(A) ${{7}}{{.0}}$
(B) ${{3}}{{.0}}$
(C) ${{5}}{{.5}}$
(D) ${{11}}$

Answer 1

In the above question, it is asked at which ${{pH}}$ , the indicator changes colour. We know that the dissociation of weak acid indicators causes the solution to change colour. So, first we have to find out the value of ${{[}}{{{H}}^{{ + }}}]$ and then we can find the ${{pH}}$ value.
Formula Used
${{pH = }} - {{log}}\left[ {{{{H}}^{{ + }}}} \right]$
Where $\left[ {{{{H}}^{{ + }}}} \right]$ = concentration of ${{{H}}^{{ + }}}$ .
${{{K}}_{{b}}}{{ = }}\dfrac{{\left[ {{{{B}}^{{ + }}}} \right]\left[ {{{O}}{{{H}}^{{ - }}}} \right]}}{{\left[ {{{BOH}}} \right]}}$
Where $\left[ {{{{B}}^{{ + }}}} \right]$ , $\left[ {{{O}}{{{H}}^{{ - }}}} \right]$ and $\left[ {{{BOH}}} \right]$ are equilibrium concentration of ${{{B}}^{{ + }}}$ , ${{O}}{{{H}}^{{ - }}}$ and ${{BOH}}$ respectively.

Complete step by step solution
Acid-base indicators are also known as ${{pH}}$ indicators. They are usually weak acids or bases which when dissolved in water dissociate slightly and form ions.
Let us first write the basic indicator equilibrium equation:
${{BOH}} \rightleftharpoons {{{B}}^{{ + }}}{{ + O}}{{{H}}^{{ - }}}$
We know that the indicator changes the colour when $\left[ {{{{B}}^{{ + }}}} \right]$ = $\left[ {{{BOH}}} \right]$ . So, the dissociation can be written as:
${{{K}}_{{b}}}{{ = }}\left[ {{{O}}{{{H}}^{{ - }}}} \right]$ = ${{1 \times 1}}{{{0}}^{{{ - 11}}}}$
Now, we have to find the value of $\left[ {{{{H}}^{{ + }}}} \right]$ . We know that:
$\left[ {{{{H}}^{{ + }}}} \right]\left[ {{{O}}{{{H}}^{{ - }}}} \right]{{ = }}{{{K}}_{{w}}}$
We know that ${{{K}}_{{w}}}$ is the dissociation constant which is equal to ${{1 \times 1}}{{{0}}^{{{ - 14}}}}$ always.
Hence, after rearranging, we get:
$\left[ {{{{H}}^{{ + }}}} \right]{{ = }}\dfrac{{{{{K}}_{{w}}}}}{{\left[ {{{O}}{{{H}}^{{ - }}}} \right]}}$
Substituting the values, we get:
$\left[ {{{{H}}^{{ + }}}} \right]{{ = }}\dfrac{{1 \times {{10}^{ - 14}}}}{{1 \times {{10}^{ - 11}}}} = {10^{ - 3}}$
Hence, we got the concentration of hydrogen ion as ${{1}}{{{0}}^{{{ - 3}}}}$ .
Now, we can find the ${{pH}}$ of the solution as:
${{pH = }} - {{log}}\left[ {{{{H}}^{{ + }}}} \right]{{ = }} - {{log(1}}{{{0}}^{ - 3}}{{)}}$
${{pH = }} - {{(}} - {{3)log(10) = 3}}$
So, ${{pH}}$ of the solution for which the indicator changes colour is 3.
Hence, the correct option is option B.

Note
For a general reaction:
${{{A}}_{{x}}}{{{B}}_{{y}}} \rightleftharpoons {{xA + yB}}$
Molecule ${{{A}}_{{x}}}{{{B}}_{{y}}}$ divided into x units of A and y units of B.
The dissociation constant can be defined as:
${{{k}}_{{d}}}{{ = }}\dfrac{{{{\left[ {{A}} \right]}^{{x}}}{{\left[ {{B}} \right]}^{{y}}}}}{{{{[}}{{{A}}_{{x}}}{{{B}}_{{y}}}{{]}}}}$
where $\left[ {{A}} \right]$ , $\left[ {{B}} \right]$ , ${{[}}{{{A}}_{{x}}}{{{B}}_{{y}}}{{]}}$ are the equilibrium concentrations of A,B and compound ${{{A}}_{{x}}}{{{B}}_{{y}}}$ .
A small dissociation constant indicates that the ligands are tightly bounded.