pH of water is 7 at $$25^{\circ}C$$.If water is heated to $$80^{\circ}C$$.,it's pH will :

We need to determine what happens to the pH of water when it is heated from $$25°C$$ to $$80°C$$.

Understand the autoionization of water.

Water undergoes autoionization: $$H_2O \rightleftharpoons H^+ + OH^-$$

The ionic product of water is: $$K_w = [H^+][OH^-]$$

At $$25°C$$, $$K_w = 1.0 \times 10^{-14}$$, so $$[H^+] = [OH^-] = 10^{-7}$$ M, giving $$\text{pH} = 7$$.

Effect of temperature on $$K_w$$.

The autoionization of water is an endothermic process ($$\Delta H > 0$$). By Le Chatelier's principle, increasing the temperature shifts the equilibrium to the right, producing more $$H^+$$ and $$OH^-$$ ions.

Therefore, at $$80°C$$, $$K_w$$ is significantly larger than $$10^{-14}$$. For example, at $$80°C$$, $$K_w \approx 2.4 \times 10^{-13}$$.

Calculate the new pH.

At $$80°C$$: $$[H^+] = [OH^-] = \sqrt{K_w} = \sqrt{2.4 \times 10^{-13}} \approx 4.9 \times 10^{-7}$$ M

$$\text{pH} = -\log(4.9 \times 10^{-7}) \approx 6.31$$

Key observations.

Although the pH decreases below 7, the water is still neutral because $$[H^+] = [OH^-]$$. The pH decreased because more ions were produced, but the solution is neither acidic nor basic.

Note that both $$[H^+]$$ and $$[OH^-]$$ increase simultaneously. Option (2) states that $$H^+$$ increases but $$OH^-$$ decreases, which is incorrect.

The correct answer is Option (1): pH decreases.