Chemistry pH Problem?

Question

What is the pH of a solution prepared by mixing 71.0 mL of 0.0298 M NaOH and 89.0 mL of 0.0263 M Ba(OH)2?
Any help would be appreciated!!

Answer 1

First off, it is important to realize that this is no trick question. Ba(OH)2 and NaOH are both strong bases. There is no buffer effect. Let the NaOH solution be called NS; let the Ba(OH)2 solution be called BS.

71.0mLNS x 0.0298molNaOH/1000mLNS = 0.00212molNaOH = 0.00212molOH-

89.0mLBS x 0.0263molBa(OH)2/1000mLBS x 2molOH-/1molBa(OH)2 = 0.00468molOH-

The moles of OH- are 0.00212 + 0.00468 = 0.00680molOH-

The volume of solution is 71.0mL + 89.0mL = 160mL.Let the combined solution be called S.

So the conentration of OH- is 0.0068molOH-/160mLS x 1000mL/1L = 0.0425mol/L. So [OH-] = 4.25x10^-2

Kw = [H+][OH-] = 10^-14 = 10 x 10^-15

[H+] = (10x10^-15)/(4.25x10^-2) =2.35x10^-13

pH = -Log[H+] = -(13 - 0.371) = 12.6