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

Is a logical series of steps