Question

The values of ${{\text{K}}_{\text{w}}}$ at the physiological temperature $\left( {{{37}^o}{\text{C}}} \right)$ is $2.56 \times {10^{ - 14}}$ . What is the pH at the neutral point of water at this temperature, where there are equal numbers of ${{\text{H}}^ + }$ and ${\text{O}}{{\text{H}}^ - }$ ions? Note: answer in nearest integer

Answer 1

Ionic product of water or ${{\text{K}}_{\text{w}}}$ is the product of hydrogen ions and hydroxide ions present in the substance. It does not depend on concentration of water and depends only on temperature.

Complete step by step answer:
For the autoprotolysis reaction of the water as follow: ${{\text{H}}_2}{\text{O}} + {{\text{H}}_2}{\text{O}} \to {\text{O}}{{\text{H}}^ - } + {{\text{H}}_3}{{\text{O}}^ + }$ , the constant can be written as ${\text{K}} = \dfrac{{\left[ {{\text{O}}{{\text{H}}^ - }} \right]\left[ {{{\text{H}}_3}{{\text{O}}^ + }} \right]}}{{{{\left[ {{{\text{H}}_2}{\text{O}}} \right]}^2}}}$ and it can be rearranged to write: ${\text{K}}.{\left[ {{{\text{H}}_2}{\text{O}}} \right]^2} = \left[ {{\text{O}}{{\text{H}}^ - }} \right]\left[ {{{\text{H}}_3}{{\text{O}}^ + }} \right]$ and this is known by the ionization constant of water or ionic product of water: ${{\text{K}}_{\text{w}}} = \left[ {{\text{O}}{{\text{H}}^ - }} \right]\left[ {{{\text{H}}_3}{{\text{O}}^ + }} \right] = \left[ {{\text{O}}{{\text{H}}^ - }} \right]\left[ {{{\text{H}}^ + }} \right]$ . The ${{\text{K}}_{\text{w}}}$ does not depend on the amount of water but depends on temperature. So, basically the ionic product of water is the product of hydrogen ions and hydroxide ions present in the solution. At ${25^o}{\text{C}}$ , the value of ${{\text{K}}_{\text{w}}}$ is ${10^{ - 14}}$ and the value of ionic product of water increases with the increasing temperature.
As given in the question, the water is at neutral point which states that concentration of hydrogen ion and hydroxide ion is equal $\left[ {{{\text{H}}^ + }} \right] = \left[ {{\text{O}}{{\text{H}}^ - }} \right]$ and the value of ${{\text{K}}_{\text{w}}}$ at the physiological temperature $\left( {{{37}^o}{\text{C}}} \right)$ is $2.56 \times {10^{ - 14}}$ , it can be given as:
${{\text{K}}_{\text{w}}} = \left[ {{{\text{H}}^ + }} \right]\left[ {{\text{O}}{{\text{H}}^ - }} \right]$ and as we know $\left[ {{{\text{H}}^ + }} \right] = \left[ {{\text{O}}{{\text{H}}^ - }} \right]$ , thereby
${{\text{K}}_{\text{w}}} = \left[ {{{\text{H}}^ + }} \right]\left[ {{{\text{H}}^ + }} \right] = {\left[ {{{\text{H}}^ + }} \right]^2}$
$\left[ {{{\text{H}}^ + }} \right] = {\left( {{{\text{K}}_{\text{w}}}} \right)^{\dfrac{1}{2}}}$ and putting the value of ${{\text{K}}_{\text{w}}}$ which is $2.56 \times {10^{ - 14}}$ ,
$\left[ {{{\text{H}}^ + }} \right] = {\left( {2.56 \times {{10}^{ - 14}}} \right)^{\dfrac{1}{2}}}$
$\left[ {{{\text{H}}^ + }} \right] = 1.6 \times {10^{ - 7}}$ . This concentration of hydrogen ions can be used to determine pH of a solution.
And as we know, pH of a solution is equal to the negative logarithm of hydrogen ion concentration. The pH can be given as: ${\text{pH}} = - {\text{log}}\left[ {{{\text{H}}^ + }} \right]$ and as we have $\left[ {{{\text{H}}^ + }} \right] = 1.6 \times {10^{ - 7}}$ , therefore pH will be equal to ${\text{pH}} = - {\text{log}}\left[ {1.6 \times {{10}^{ - 7}}} \right] = 7 - 0.20 = 6.8$ .

Thus, the correct answer is $6.8$ or 7.

Note:
The value of ${{\text{K}}_{\text{w}}}$ does not depend upon the nature of mixture whether it is acidic, basic or neutral. It is constant for a given temperature condition.