If one were to mix 200.0 mL of an HCl solution that had a pH of 3.00 with 200.0 mL of an NaOH solution that had a pH of 10.00, what would be the pH of the final solution?
Can you show me the steps please so that I can apply to questions that ask for the same type of solving? thank you!
pH = -log (H^+)
pH = 3.00, (H^+) = 1E-3
pH = 10.00
(H^+) = 1E-10 and OH^- = 1E-4
moles = M x L
For HCl = 1E-3 x 0.200 = 2E-4 moles.
For NaOH = 1E-4 x 0.200 = 2E-5 moles.
.............HCl + NaOH ==> NaCl + H2O
initial..... 2E-4..2E-5.......0....0
change......-2E-5.-2E-5....+2E-5 .+2E-5
final.......1.8E-4...0......2E-5..2E-5
So the final solution contains an excess of 1.8E-4 moles HCl, no NaOH, and NaCl and H2O formed don't change the pH, in a final volume of 400 mL (0.400 L).
(H^+) = moles/L = 1.8E-4/0.400 = 4.5E-4
pH = -log(H^+) = -log(4.5E-4) = -(3.35) = 3.35.