Question

If ionic product of water is \[{K_w}{\text{ }} = {\text{ }}{10^{ - 16}}\] at \[4^\circ {\text{C}}\] , then a solution with \[{\text{pH = 7}}{\text{.5}}\] at \[4^\circ {\text{C}}\] will:
A. Turn blue litmus red
B. Turn red litmus blue
C. Turn turmeric paper brown
D. Be neutral to litmus

Answer 1

Hint:The ionic product of water represents equilibrium constant expression for the autoionization of water. You can use this expression to obtain hydronium ion concentration in a neutral solution. From hydronium ion concentration, you can calculate the pH of the solution. You can compare the given value of pH with pH of neutral solution and say if the solution is acidic or alkaline. The colour change of litmus paper is different for acidic and alkaline solutions.

Complete answer:
At \[4^\circ {\text{C}}\] the ionic product of water is \[{K_w}{\text{ }} = {\text{ }}{10^{ - 14}}\]
The ionic product of water is the product of hydronium ion concentration and hydroxide ion concentration. Hence, at room temperature,
\[{K_w}{\text{ }} = {\text{ }}{10^{ - 14}} = \left[ {{{\text{H}}_3}{{\text{O}}^ + }} \right] \times \left[ {{\text{O}}{{\text{H}}^ - }} \right]\]
For neutral water, the hydronium ion concentration is equal to the hydroxide ion concentration.
\[\left[ {{{\text{H}}_3}{{\text{O}}^ + }} \right] = \left[ {{\text{O}}{{\text{H}}^ - }} \right]\]
Hence, \[{K_w}{\text{ }} = {\text{ }}{10^{ - 14}} = \left[ {{{\text{H}}_3}{{\text{O}}^ + }} \right] \times \left[ {{{\text{H}}_3}{{\text{O}}^ + }} \right] = {\left[ {{{\text{H}}_3}{{\text{O}}^ + }} \right]^2}\]
Take square root on both sides of the equation
\[\Rightarrow \left[ {{{\text{H}}_3}{{\text{O}}^ + }} \right] = \sqrt {{K_w}} {\text{ }} = {\text{ }}\sqrt {{{10}^{ - 14}}} = {10^{ - 7}}{\text{ M}}\]
Calculate the pH of neutral water at room temperature
\[\Rightarrow pH = - {\log _{10}}\left[ {{{\text{H}}_3}{{\text{O}}^ + }} \right] = - {\log _{10}}{10^{ - 7}}{\text{ M = 7}}\]
Acidic solutions have pH value less than 7 at room temperature and alkaline solutions have pH value greater than 7 at room temperature.
Now consider the situation in which the temperature is reduced to \[4^\circ {\text{C}}\] .
At \[4^\circ {\text{C}}\] the ionic product of water is \[{K_w}{\text{ }} = {\text{ }}{10^{ - 16}}\]
\[\Rightarrow {K_w}{\text{ }} = {\text{ }}{10^{ - 16}} = \left[ {{{\text{H}}_3}{{\text{O}}^ + }} \right] \times \left[ {{{\text{H}}_3}{{\text{O}}^ + }} \right] = {\left[ {{{\text{H}}_3}{{\text{O}}^ + }} \right]^2}\]
Take square root on both sides of the equation
\[\Rightarrow \left[ {{{\text{H}}_3}{{\text{O}}^ + }} \right] = \sqrt {{K_w}} {\text{ }} = {\text{ }}\sqrt {{{10}^{ - 16}}} = {10^{ - 8}}{\text{ M}}\]
Calculate the pH of neutral water at \[4^\circ {\text{C}}\]
\[\Rightarrow pH = - {\log _{10}}\left[ {{{\text{H}}_3}{{\text{O}}^ + }} \right] = - {\log _{10}}{10^{ - 8}}{\text{ M = 8}}\]
Neutral solutions have a pH value of 8 at \[4^\circ {\text{C}}\] . Acidic solutions have pH value less than 8 at \[4^\circ {\text{C}}\] and alkaline solutions have pH value greater than 8 at \[4^\circ {\text{C}}\] .
Then a solution with \[{\text{pH = 7}}{\text{.5}}\] at \[4^\circ {\text{C}}\] will be acidic in nature as it has pH value less than 8:
An acidic solution turns blue litmus red.
Hence, a solution with \[{\text{pH = 7}}{\text{.5}}\] at \[4^\circ {\text{C}}\] will turn blue litmus red.

Hence, the correct option is the option A.

Note:
A neutral solution has no effect on the litmus paper. In other words, a neutral solution will not change the colour of either blue litmus paper or red litmus paper. An acidic solution changes the colour of blue litmus paper to red. An alkaline solution changes the colour of red litmus paper to blue.