[webassignbuddy]

What is the pH (to nearest 0.01 pH unit) of a solution prepared by mixing 64.0 mL of 0.0633 M NaOH and 79.0 mL of 0.0322 M Ba(OH)2?

Step 1) Determine the limiting reagent

mmol NaOH: 64 mL x (0.0633 mmol/mL) = 4.0512 mmol OH- (Base)
mmol Ba(OH)2: 79 mL x (0.0322 mmol/mL) = 2.5438 mmol x 2 = 5.0876 mmol OH- (Base)

UPDATE: Wait nevermind! I figured it out! I can disregard the table below.

Step 2)

[OH-] = total mmol OH- / total mL
[OH-] = (4.0512+5.0876)/(69+39)
[OH-] = 0.0639 M

Step 3) Solve for pH of the solution

pOH = -log[OH-]
pOH = -log(0.0639)
pOH = 1.194499

pH = 14 - pOH
pH = 14 - 1.194499
pH = 12.8



I got it right this time.

And I just realized that Borek so that's why I had to make the update! 

[Borek]

Since when strong base reacts as an acid?

What happens when you mix two bases? Or two acids?

Why do you think there is a reaction taking place?