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

Hint: 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.