At 65°c, the ion-product constant of water, kw, is 1.20 ´ 10-13. the ph of pure water at 65°c is:

The pH of the pure water in the given solution is 6.46.

The given parameters;

temperature of the water, t = 65 ⁰Cequilibrium constant of the water, kw = 1.2 x 10⁻¹³

The concentration of water consist hydroxyl ion and hydrogen ion;

For pure water we can assume that the water contains only hydrogen ion;

The ph of the pure water concentration is calculated as follows;

Thus, the pH of the pure water in the given solution is 6.46.

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When Kw = [OH-] [H+] & when we know that the pure water should has the concentration of H [ H+] = the concentration of OH [OH-]So
[H+]=[OH-]
∴ we can assume that Kw = [H+] [H+] or Kw = [OH-] [ OH-] or we can assume [H+] or [ OH-] as z
So, then Kw = Z^2

when we have the value of Kw so, by substitution:
Z^2 = 1.2 x 10 ^-13
Z = √ 1.2x10^-13
   =  3.46x 10^-7 
So we now get [H+] = 3.46X10^-7 
when PH = -㏒[H+]
∴ PH = -㏒(3.46x10^-7) = 6.46

Ibelieve the answer is d

Idon't see the pic?