Question

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

Answer 1

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

Answer 2

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

The given parameters;

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

The concentration of water consist hydroxyl ion and hydrogen ion;

$kw = [OH^{-1}][H^{+}]$

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

$kw = [H^{+}][H^{+}]\\\\kw = [H^{+}]^2\\\\1.2\times 10^{-13} = [H^{+}]^2\\\\\sqrt{(1.2\times 10^{-13} )} = [H^{+}]\\\\3.46 \times 10^{-7} = [H ^{+}]$

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

$pH = -log[H^{+}]\\\\pH = -log [3.46\times 10^{-7}]\\\\pH = 6.46$

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

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