1. Auto-ionization of water

   Water undergoes self-ionization: $2H_2O \rightleftharpoons H_3O^+ + OH^-$ or simply, $H_2O \rightleftharpoons H^+ + OH^-$

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

2. At $25^\circ C$

   At $25^\circ C$, $K_w = 1.0 \times 10^{-14}$ For pure water, $[H^+] = [OH^-] = 1.0 \times 10^{-7}$ Therefore, $\text{pH} = -\log(10^{-7}) = 7$

3. Effect of increasing temperature

   The ionization of water is an endothermic process. Hence, when temperature increases, equilibrium shifts forward and $K_w$ increases.

   So at $80^\circ C$: $K_w > 10^{-14}$ Therefore in pure water, $[H^+] = [OH^-] = \sqrt{K_w}$ and both increase.

4. Effect on pH

   Since $[H^+]$ increases, $\text{pH} = -\log[H^+]$ will decrease.

   Note: Even though pH becomes less than 7, the water is still neutral because: $[H^+] = [OH^-]$

5. Checking options

   • A: Decrease — Correct
   • B: Remains the same — Incorrect
   • C: Increase — Incorrect
   • D: $H^+$ concentration increases, $OH^-$ concentration decreases — Incorrect, because both increase equally in pure water

Final Answer: $\boxed{\text{A: Decrease}}$