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# For pure water:(A) pH increases while pOH decreases with rise in temperature(B) pH decreases while pOH increases with rise in temperature(C) Both pH and pOH decreases with rise in temperature (D) Both pH and pOH increases with rise in temperature

Hint: pH and pOH are temperature dependent. When the temperature rises, the rate of ionization also changes accordingly. Similarly when the temperature will fall, there will be a change in both the pH and pOH values for pure water.
For pure water, at $25^\circ$c,$\left[ {{H^ + }} \right] = {10^{ - 7}}$ and $\left[ {O{H^ - }} \right] = {10^{ - 7}}$

At a temperature of $100^\circ$c, ${k_w}$ (that is ionic product of water) of Boiling Water is ${10^{ - 12}}$, then we know that
$\left[ {{H^ + }} \right]\left[ {O{H^ - }} \right] = {k_w}$
$\left[ {{H^ + }} \right]\left[ {O{H^ - }} \right] = {10^{ - 12}}$
We clearly know that $\left[ {{H^ + }} \right] = \left[ {O{H^ - }} \right]$
$\Rightarrow$${\left[ {{H^ + }} \right]^2} = {10^{ - 12}}$
$\Rightarrow$$\left[ {{H^ + }} \right] = {10^{ - 6.36}}$
$\Rightarrow$$pH = - log\left[ {{H^ + }} \right] = - \log \left( {{{10}^{ - 6}}} \right)$
$\Rightarrow$$pH = 6$
At a temperature of $50^\circ$c, $p{K_w} = 13\cdot36$, (where $p{K_w}$ is the negative log of ${k_w}$) then we also know that
$pH + pOH = p{K_w}$
We are aware of the fact that $pH = pOH$
$\Rightarrow$$pH = \dfrac{{13.36}}{2} = 6\cdot68$
So from the above whole discussion, we can conclude that when the temperature of pure water decreases, pH and pOH values of both will increase.
Or we can also say that pH and pOH values will decrease when the temperature of the pure water will increase.

The Correct Answer of the above question is option C. both pH and pOH decreases with rise in temperature.

Note: pH and pOH values are always the same for pure water. And both are temperature dependent. When the temperature rises, both values of pH and pOH decrease, while these values increase when the temperature of pure water falls down.