Question

If \[{K_W}\] for water is \[9.62 \times {10^{ - 14}}\] at \[{60^\circ }C\]. Calculate pH of water.

Answer 1

Hint: Dissociation constant (or) Ionic product of water \[{K_W}\] is defined as the product of the molar concentration of hydroxyl ion and hydrogen ion concentration at a constant temperature. For pure water, hydroxyl ion and hydrogen ion concentration must be equal.

Complete step by step answer:
\[{K_W}\] is defined as the product of the molar concentration of hydroxyl ion and hydrogen ion concentration at a constant temperature. It can be written as follows,
\[\left[ {{H^ + }} \right]\left[ {O{H^ - }} \right] = {K_W}\]
It is known that for pure water, hydrogen ion concentration should be equal to hydroxyl ion concentration.
\[\left[ {{H^ + }} \right] = \left[ {O{H^ - }} \right]\]
\[ \Rightarrow {\left[ {{H^ + }} \right]^2} = {K_W}\]
Thus, for pure water at a constant temperature, hydrogen ion concentration can be obtained as the square root of the ionic product of water.
\[ \Rightarrow {\left[ {{H^ + }} \right]^2} = 9.62 \times {10^{ - 14}}\]
At\[{60^\circ }C\], \[ \Rightarrow \left[ {{H^ + }} \right] = \sqrt {9.62 \times {{10}^{ - 14}}} \]
\[ \Rightarrow \left[ {{H^ + }} \right] = 3.102 \times {10^{ - 7}}\]
It is known that pH is the negative logarithm of the concentration of free hydrogen.
\[pH = - \log \left[ {{H^ + }} \right]\]
\[ \Rightarrow pH = - \log \left[ {3.102 \times {{10}^{ - 7}}} \right]\]
\[pH = 6.50\]

Thus, pH for water is 6.50.

Note: It is known that pH is the measure of hydrogen ion concentration and pOH is the hydroxyl ion concentration.
Thus, the ionic product of water can be explained as similar to hydrogen ion concentration can be written as,
\[p{K_W} = - \log \left[ {{K_W}} \right]\]
But we know that the Ionic product of water \[{K_W}\] is defined as the product of the molar concentration of hydroxyl ion and hydrogen ion concentration at a constant temperature.
Thus, \[p{K_W} = - \log \left[ {{H^ + }} \right]\left[ {O{H^ - }} \right]\]
\[ \Rightarrow p{K_W} = - \log \left[ {{H^ + }} \right] - \log \left[ {O{H^ - }} \right]\]
\[ \Rightarrow p{K_W} = pH + pOH\]
Thus, \[p{K_W}\] is the sum of pH and pOH. Thus, if one of concentration{hydrogen ion /hydroxyl ion} is known, then \[{K_W}\] of water is calculated since for pure water, hydroxyl ion and hydrogen ion concentration must be equal. And also if pH/pOH is known, then \[p{K_W}\] the value can be calculated.