What is the pH of a solution prepared by mixing 65.0 mL of 0.0686 M NaOH and 79.0 mL of 0.0218 M Ba(OH)2?

The dissociation of each mole of NaOH gives 1 mole of OH⁻ ions, while the dissociation of each mole of Ba(OH)₂ gives 2 moles of OH⁻ ions.

Total moles of OH⁻ in the solution
= (0.0686 mol/L) × (65.0/1000 L) + (0.0218 mol/L) × (79.0/1000 L) × 2
= 0.007903 mol

Assuming the volume of solution is additive, total volume of the final solution
= (65.0 + 79.0) mL
= 144 mL
= 0.144 L

[OH⁻]
= (0.007903 mol) / (0.144 L)
= 0.05488 M

pH
= 14.00 - pOH
= 14.00 + log(0.05488)
= 12.74

moles OH- from NaOH = 0.0650 L X 0.0686 mol/L = 4.46X10^-3 mol OH-
Moles OH- from Ba(OH)2 = 0.0790 L X 2(0.0218 mol/L) = 3.44X10^-3 mol OH-

Total moles OH- = 7.90X10^-3 mol
Total volume = 144 mL = 0.144 L

Molarity OH- = 7.90X10^-3 / 0.144 L = 0.0549 M
pOH = 1.26
pH = 14.00 - 1.26 = 12.74