At 65°C,the ion-product constant of water,K_w,is 1.20 $\times$ 10^-¹³.The pH of pure water at 65°C is: A)7.000 B)6.560 C)5.880 D)6.460 E)none of these

Question

Quizplus · Accepted Answer

At a specific temperature, the ion product constant of water is given by:
KW = [H+] [OH-]
At 65°C, KW = 1.20 x 10^-10.
We know that in pure water, [H+] = [OH-], so we can say that [H+] = [OH-] = x.
Substitute this in the KW equation:
KW = x^2
1.20 x 10^-10 = x^2
x = 1.095 x 10^-5 M
pH = -log[H+]
pH = -log(1.095 × 10^-5) = 6.462

Therefore, the pH of pure water at 65°C is 6.462, which is option D.

At 65°C,the Ion-Product Constant of Water,Kw,is 1 ×\times× 10-13

At 65°C,the Ion-Product Constant of Water,K_w,is 1 $\times$ 10^-¹³