Question

At some high temperature, ${{\text{K}}_{\text{w}}}$of water is ${{10}^{-13}}$. Then the $\text{pH}$ of the water at the same temperature is:
(A) $8.3$
(B) $\text{6}\text{.5}$
(C) $7.42$
(D) $6$

Answer 1

$\text{pH}$ is the negative logarithm of hydrogen ion concentration. It is the measure of acidic and basic nature of the solution.
On increasing temperature, both ${{\text{H}}^{\text{+}}}$ and $\text{O}{{\text{H}}^{\text{-}}}$concentration increases equally so, the water remain neutral but natural $\text{pH}$ change from $7$ to $6$ at ${{90}^{\circ }}C$.

Complete answer:
${{\text{K}}_{\text{w}}}$ is called an ionic product or dissociation constant of water. It is equal to the product of $\text{ }\\!\\![\\!\\!\text{ }{{\text{H}}^{\text{+}}}]$ $\text{ }\\!\\![\\!\\!\text{ O}{{\text{H}}^{\text{-}}}]$. At constant temperature of${{25}^{\circ }}C$, the value of ${{\text{K}}_{\text{w}}}$ is $1\times {{10}^{-14}}$.
${{\text{H}}_{\text{2}}}\text{O}\,\,\rightleftharpoons \,\,{{\text{H}}^{\text{+}}}\,\text{+}\,\,\text{O}{{\text{H}}^{\text{-}}}$
${{\text{K}}_{\text{w}}}\text{=}\,\text{ }\\!\\![\\!\\!\text{ }{{\text{H}}^{\text{+}}}\text{ }\\!\\!]\\!\\!\text{ }\,\text{ }\\!\\![\\!\\!\text{ O}{{\text{H}}^{\text{-}}}\text{ }\\!\\!]\\!\\!\text{ }....\text{(i)}$
..Ionization of water is an endothermic process, so increasing temperature degree of dissociation increases or both and concentration increases.
So, at high temperature ${{\text{K}}_{\text{w}}}=\,{{10}^{-13}}$ is given
At any temperature $\text{ }\\!\\![\\!\\!\text{ }{{\text{H}}^{\text{+}}}]$=$\text{ }\\!\\![\\!\\!\text{ O}{{\text{H}}^{\text{-}}}]$because nature of water is neutral.

& \text{p}{{\text{K}}_{\text{w}}}\text{=}\,\text{-log }\\!\\![\\!\\!\text{ }{{\text{K}}_{\text{w}}}\text{ }\\!\\!]\\!\\!\text{ } \\\ & \text{p}{{\text{K}}_{w}}=\,-\log [1\times {{10}^{-13}}] \\\ & \text{p}{{\text{K}}_{\text{w}}}\text{=}\,\text{13} \\\ \end{aligned}$$ Thus from the equation … (i) $$\begin{aligned} & {{\text{K}}_{\text{w}}}\text{=}\,\text{ }\\!\\![\\!\\!\text{ }{{\text{H}}^{\text{+}}}\text{ }\\!\\!]\\!\\!\text{ }\,\text{ }\\!\\![\\!\\!\text{ O}{{\text{H}}^{\text{-}}}\text{ }\\!\\!]\\!\\!\text{ }....\text{(i)} \\\ & \text{after}\,\text{taking}\,\text{-}\,\text{log}\,\text{in}\,\text{both}\,\text{side} \\\ & \text{-log }{{\text{K}}_{\text{w}}}\text{=}\,\text{-log }\\!\\![\\!\\!\text{ }{{\text{H}}^{\text{+}}}\text{ }\\!\\!]\\!\\!\text{ }\,\text{-log }\\!\\![\\!\\!\text{ O}{{\text{H}}^{\text{-}}}\text{ }\\!\\!]\\!\\!\text{ }\,\, \\\ & \text{p}{{\text{K}}_{\text{w}}}\,\text{=}\,\text{pH}\,\text{+}\,\text{pOH} \\\ & \text{p}{{\text{K}}_{\text{w}}}\,\text{=}\,\text{2pH}\,\,\,\,\text{ }\\!\\!\\{\\!\\!\text{ because}\,\text{for}\,\text{water}\,\text{pH=}\,\,\text{pOH }\\!\\!\\}\\!\\!\text{ } \\\ & \text{pH}\,\text{=}\,\frac{\text{p}{{\text{K}}_{\text{w}}}}{\text{2}} \\\ & \text{pH}\,\,\text{=}\,\frac{\text{13}}{\text{2}} \\\ & \text{pH}\,\,\text{=}\,\text{6}\text{.5} \\\ \end{aligned}$$ **So the option (B) will be the correct answer.** **Additional information:** When an acid is added to water ${{\text{H}}^{\text{+}}}$ ion from acid combines with the $\text{O}{{\text{H}}^{\text{-}}}$ ions so that $${{\text{K}}_{\text{w}}}$$ remains constant. Thus addition of an acid decreases the conc. Of $\text{O}{{\text{H}}^{\text{-}}}$ ion and addition of base decreases the conc. of ${{\text{H}}^{\text{+}}}$ ions. In both cases the self-ionisation of water is suppressed due to the extra supply of ${H^+}$ and ${OH^-}$ ions. **Note:** Water is a very weak electrolyte, so ionisation of water increases with increasing the temperature. The addition of salt, acid or base does not change the value of$${{\text{K}}_{\text{w}}}$$, its value changes with temperature only. On increasing temperature $\text{pH}$ value always decreases.