In the process of water treatment at 298°K, the kw of 1014 mol2 dm-6 as the temperature of the system changed to 300°K. What is the pH at 300° K?

 

The value given, ${�}_{�}=10^{-14}{\text{mol}}^2{\text{dm}}^{-6}$, represents the ion product of water at 298 K. The ion product of water, ${�}_{�}$, is given by the expression:

${�}_{�}=[{\text{H}}^{+}][{\text{OH}}^{-}]$

At 298 K, ${�}_{�}=10^{-14}{\text{mol}}^2{\text{dm}}^{-6}$. At 300 K, we want to find the new concentration of ${\text{H}}^{+}$ ions ($[{�}^{+}]$) and ${\text{OH}}^{-}$ ions ($[�{�}^{-}]$) and then calculate the pH.

Since water is neutral, $[{�}^{+}]=[�{�}^{-}]$. Let's denote this common value as $�$.

At 300 K, ${�}_{�}$ will change due to the change in temperature. The new ${�}_{�}$ at 300 K will be:

${�}_{�}^{\mathrm{\prime }}=10^{-14+\mathrm{\Delta }�}=10^{-14+(300-298)}=10^{-14+2}=10^{-12}{\text{mol}}^2{\text{dm}}^{-6}$

Now, ${�}_{�}^{\mathrm{\prime }}=[{�}^{+}][�{�}^{-}]$, which is equal to ${�}^2$.

So, ${�}^2=10^{-12}$.

Taking the square root of both sides:

$�=\sqrt{10^{-12}}=10^{-6}\text{mol}{\text{dm}}^{-3}$

This is the concentration of both ${\text{H}}^{+}$ and ${\text{OH}}^{-}$ ions in the solution at 300 K.

Now, to find the pH, we use the formula:

$\text{pH}=-{\log }_{10}[{�}^{+}]$

$\text{pH}=-{\log }_{10}(10^{-6})$

$\text{pH}=-(-6)$

$\text{pH}=6$

So, the pH of the solution at 300 K is 6.