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

<p>When water is heated, its self-ionization increases. At 25°C, the water ionization constant is given by</p> <p>$K_w = [\mathrm{H}^+][\mathrm{OH}^-] = 1.0 \times 10^{-14},$</p> <p>so in pure water, </p> <p>$[\mathrm{H}^+] = [\mathrm{OH}^-] = 1.0 \times 10^{-7} \, \text{M},$</p> <p>which corresponds to a pH of 7.</p> <p>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.</p> <p>Key points:</p> <p><p>Heating water increases $K_w$.</p></p> <p><p>Higher $K_w$ means higher $[\mathrm{H}^+]$ (and equally higher $[\mathrm{OH}^-]$).</p></p> <p><p>Thus, the pH decreases (e.g., it might be around 6.5–6.6 at 80°C) even though the water is still neutral.</p></p> <p>So, the correct answer is:</p> <p>Option A – Decrease.</p>