If the pOH of a solution is 11.6 the [H+] would be

Answer 1

Hello, Sammy,

pOH and pH are the negative logs of the OH and H concentrations. So a pOH of 11.6 would mean that the concentration of OH is

[OH^-] = 1x10^-11.6 M

At STP, pure water is in equilibrium with it's component ions:

H₂O = H⁺ + OH^-

The equilibrium coefficient for this is 1x10^-14. So we can write

1x10^-14 =[H⁺]*[OH^-]

Now we can find [H⁺]

1x10^-14 =[H⁺]*[1x10^-11.6]

[H⁺] = [1x10^-11.6]/[1x10^-14]

[H⁺] = [1x10^+2.4]

The negative log of this number is 2.4

pH = 2.4

The above calculation is the fundamental relationship between pH and pOH of water. Once this is understood, it makes sense to use a much quicker approach. pH and pOH must always sum to 14 for water. If pOH is 11.6, then pH is 14 - 11.6, or 2.4.

Bob