JEE Main 2025 — Ionic Equilibrium Question with Solution

When water is heated, its self-ionization increases. At 25°C, the water ionization constant is given by

$$K_w=[{\mathrm{H}}^{+}][{\mathrm{OH}}^{-}]=1.0\times 10^{-14},$$

so in pure water,

$$[{\mathrm{H}}^{+}]=[{\mathrm{OH}}^{-}]=1.0\times 10^{-7}\text{M},$$

which corresponds to a pH of 7.

As water is heated to 80°C, the value of $$K_w$$ increases due to the endothermic nature of water’s autoionization. This means that both $$[{\mathrm{H}}^{+}]$$ and $$[{\mathrm{OH}}^{-}]$$ increase. However, since their concentrations remain equal (which defines neutrality at that temperature), the pH isn’t 7 anymore. Instead, the increased concentration of $${\mathrm{H}}^{+}$$ ions leads to a pH value lower than 7.

Key points:

Heating water increases $$K_w$$.

Higher $$K_w$$ means higher $$[{\mathrm{H}}^{+}]$$ (and equally higher $$[{\mathrm{OH}}^{-}]$$).

Thus, the pH decreases (e.g., it might be around 6.5–6.6 at 80°C) even though the water is still neutral.

So, the correct answer is:

Option A – Decrease.