H2O(l)+H2O(l) ⇄ H3O+(aq)+OH−(aq)

Question

At 5.0°C, the value of Kw for the equilibrium shown above is 1.9×10−15 and the value of pKw is 14.73. Which of the following is correct for pure water at this temperature?
A. [H3O+] = √1.9×10^(−15)
B. pH = −log(1.9×10^(−15))
C. 14.73 = [H3O+]eq [OH−]eq
D. pOH = pH+14.73

Answer 1

A is correct.

Answer 2

B. pH = −log(1.9×10^(−15))
C. 14.73 = [H3O+]eq [OH−]eq
D. pOH = pH+14.73
B is incorrect. To be correct it should be pH = - log(H3O+)/
C is incorrect. To be correct it should be 14.73 = pH + pOH or
Kw = (H3O^+)(OH^-)
D is incorrect. To be correct it should be pOH + OH = 14.73 = pKw

Answer 3

Thanks DrBob222!