Question

All of the above are conjugale LLIU-udse palls. 13. a. K= [HE] What is the pH of pure water at 40.0°C if the Kw at this temperature is 2.92 x 10°*? 6.767 H₂O = H+ + OH 0.465 7.000 7.233 8.446 6.767=-109x 107.233 = x X=5.8x10 -1767 = x

Answer 1

We know the formulas

Kw = [H⁺][OH^-]

pH = -log[H⁺]

We know that the pure water is always neutral and has the pH = 7 and concentration of [H⁺] is equal to [OH^-]

so for neutral water [H⁺]=[OH^-]

hence

Kw = [H⁺][H⁺] = [H⁺]²

hence

[H⁺]² = 2.92 $\times$ 10^-14

by taking square roots on bot sides we get

[H⁺] = 1.71 $\times$ 10^-7  

now

pH = -log(1.70 $\times$ 10^-7)

= 6.767

hence our answer is

pH = 7.767 Option (a) is correct