Question

The value of ${{\text{K}}_{\text{w}}}$ at $323{\text{ K}}$ is $5.474 \times {10^{ - 14}}$. $5.474 \times {10^{ - 14}}$Calculate the pH of water.

Answer 1

Hint:To solve this we must know the expression for ${{\text{K}}_{\text{w}}}$ of water. From the expression of ${{\text{K}}_{\text{w}}}$, calculate the hydrogen ion concentration. Then calculate the pH from the hydrogen ion concentration.

Formulae Used:
${{\text{K}}_{\text{w}}} = [{{\text{H}}^ + }][{\text{O}}{{\text{H}}^ - }]$
${\text{pH}} = - \log [{{\text{H}}^ + }]$

 Complete step by step answer:
We know that water dissociates into the hydrogen ion $[{{\text{H}}^ + }]$ and the hydroxide ion $[{\text{O}}{{\text{H}}^ - }]$. The dissociation reaction of water is as follows:
${{\text{H}}_2}{\text{O}} \rightleftharpoons {{\text{H}}^ + } + {\text{O}}{{\text{H}}^ - }$
We know that the ionic product of water $\left( {{{\text{K}}_{\text{w}}}} \right)$ is the product of concentration of the hydrogen ion $[{{\text{H}}^ + }]$ and the hydroxide ion $[{\text{O}}{{\text{H}}^ - }]$. Thus,
${{\text{K}}_{\text{w}}} = [{{\text{H}}^ + }][{\text{O}}{{\text{H}}^ - }]$
Where ${{\text{K}}_{\text{w}}}$ is the ionic product of water.
We are given that the value of ${{\text{K}}_{\text{w}}}$ at $323{\text{ K}}$ is . Thus,
$5.474 \times {10^{ - 14}} = [{{\text{H}}^ + }][{\text{O}}{{\text{H}}^ - }]$
From the reaction, we can see that water dissociates to form equal amounts of hydrogen and hydroxide ions. Thus, the number of hydrogen ion is equal to the number of hydroxide ions. Thus,$2.339 \times {10^{ - 7}}$
$[{{\text{H}}^ + }] = [{\text{O}}{{\text{H}}^ - }]$
Thus,
${[{{\text{H}}^ + }]^2} = 5.474 \times {10^{ - 14}}$
$[{{\text{H}}^ + }] = \sqrt {5.474 \times {{10}^{ - 14}}} $
$[{{\text{H}}^ + }] = 2.339 \times {10^{ - 7}}$
Thus, the hydrogen ion concentration is .
Now, we know that the pH is the negative logarithm of the hydrogen ion concentration. Thus, the expression for pH is as follows:
${\text{pH}} = - \log [{{\text{H}}^ + }]$
Substitute $2.339 \times {10^{ - 7}}$ for the hydrogen ion concentration. Thus,
${\text{pH}} = - \log \left( {2.339 \times {{10}^{ - 7}}} \right)$
${\text{pH}} = - \left( { - 6.631} \right)$
${\text{pH}} = 6.6$
Thus, the pH of water is 6.6.

Note: If the pH of the solution is less than 7 then the solution has acidic nature. If the pH of the solution is more than 7 then the solution has basic nature. If the pH of the solution is equal to 7 then the solution is neither acidic nor basic it is neutral in nature. The pH of water is 6.6 and thus, the water is slightly acidic.