For each of the following strong base solutions, determine [OH-],[H3O+], pH, and pOH of 1.13*10^-2 M Ba (OH) 2, 1.8x10^-4 M KOH, and 5.3x10^-4 M Ca (OH) 2

PH calculations of strong base solutions are pretty direct. It all depends on how many OH - ions are dissociated and the concentration of the solution.

Ba (OH) 2:

Since there are 2 OH - ions, the concentration of [OH-] is twice the concentration of the solution. Take the pOH by using the equation pOH = - log[OH-]. From here, you can get the pH by subtracting the pOH from 14. Finally, calculate the [H3O+] or [H+] concentration by using the equation pH = - log[H+] or [H+] = 10^ (-pH)

[OH-] = 2 x 1.13x10^-2 M = 2.26x10^-2

pOH = - log[OH-] = 1.65

pH = 14 - 1.65 = 12.35

[H3O+] = 10^ (-12.35) = 4.47x10^-13

Follow the same rule for the other compounds.

KOH:

[OH-] = 1.8x10^-4 M

pOH = - log[OH-] = 3.74

pH = 14 - 3.74 = 10.26

[H3O+] = 10^ (-10.26) = 5.50x10^-11

Ca (OH) 2:

[OH-] = 2 x 5.3x10^-4 M = 1.06x10^-3 M

pOH = - log[OH-] = 2.97

pH = 14 - 2.97 = 11.03

[H3O+] = 10^ (-11.03) = 9.33x10^-12