Question

The ionic product of water does not change with the temperature - True or false?

• A. True
• B. False

Answer 1

The correct option is B

False

$H_{2}O$ (l) + $H_{2}O$ (l) $\leftrightharpoons$ ${H_{3}O}^{+}$ (aq) + ${OH}^{-}$ (aq)
In the above auto-protolysis or self ionization reaction of water, $K_w$ = $K_{ionization}$ [$H_{2}O$] = $K_{eq}$ [$H_{2}O$] [$H_{2}O$]
You should all be able to derive $K_{eq}$.
In that reaction, the concentration of water practically does not change throughout the reaction. Hence it makes sense to go further than $K_{ionization}$ to get a new constant called the ionic product of water, which is defined as
$K_w$ = [${H_{3}O}^{+}$] [${OH}^{-}$] = 1 x ${10}^{-14}$ ${\left({MolL}^{-1}\right)}^2$
The value is obtained experimentally. And it is STRICTLY applicable only for 298 K
Another approach to deriving the same is via
$H_{2}O$(l) $\leftrightharpoons$ $H^{+}$ (aq) + ${OH}^{-}$(aq)
$K_{ionization}$ = $\frac{\left[H^{+}\right]\left[OH\right]}{H_{2}O})etc.[H^{+}$] is regarded the same as [${H_{3}O}^{+}$] and are used interchangeably.
pH is a convenient method to express the concentration of H+ ions. It is defined as pH = -log[$H^{+}$]
where [$H^{+}$] is measured in molarity or mol.$L^{-}$
pH = -log[${OH}^{-}$]; where [$H^{+}$] is measured in molarity or mol.$L^{-}$

From the ionic product it wollows that at 298 K pH = pOH = 7 (only at 298 K)