Answer:
See Explanation
Step-by-step explanation:
Let us recall that;
[H^+] [OH^-] = 1 * 10^-14
Where;
[OH^-] = 1 x 10^-11
Then;
[H^+] = 1 * 10^-14/1 x 10^-11
[H^+] = 1 * 10^-3
pH = -log [H^+]
pH = -log(1 * 10^-3)
pH = 3
Also;
pH + pOH = 14
pOH = 14 - pH
pOH = 14 -3
pOH = 11