Flask A contains 100 milliliters of a solution with a pH of 5. After you add 100 milliliters of a solution from Flask B, the pH rises to 7. What was the pH of the solution in Flask B?

Answer 1

I would do this.
pH = 5 means (H^+) - 1E-5 M.
mols (H^+) in flask A = M x L = 1E-5*0.1 = 1E-6. So to make a pH = 7 we must exactly neutralize the acid to give pure H2O which has a pH of 7.00. So we need exactly 1E-6 mols OH^- in flask B and that in 100 mL (0.1L) is (OH^-) = 1E-6/0.1 = 1E-5M OH^- which has a pOH = 5. Since pH + pOH = pKw = 14, then pH must be 9.

I worried about the dilution factor (100 mL becoming 200 mL final volume) and how to adjust for that until I finally realized that we are using H^+ to neutralize OH^- which makes H2O and adding 100 mL H2O to another 100 mL H2O just gives 200 mL of H2O but still with a pH of 7. For whatever it's worth, when you've taken more chemistry you will find that activity coefficients will change the pH of water with a salt in it to less than pH 7. I have ignored any effects due to activity instead of concn. That's makes it MUCH more complicated.