191.96mL of 0.088M barium hydroxide are titrated with 0.007M HCl?

Question

What is the pH of the solution after the addition of 12.65mL of acid solution

Answer 1

That's pretty messy. Lets try a simpler version.
We have 200 mL of 0.1 M Ba(OH)2 solution titrated with 0.01 M HCl, and 10 ml of acid solution is added

Chemistry H+ + OH- -> H2O
Barium is an alkaline earth element, so it is expected to totally dissociate and all OH- is in solution. Each mole of the hydroxide supplies 2 moles of hydroxyl ion. We have 0.2 L x 0.2 N OH- or
a total of 0.04 moles of OH-. The acid supplies 0.0001M of H+ . From the above reaction, there are 0.0399 moles of OH- left. But we have 210 mL of solution. So, the [OH-]= 0.0399 moles/0.210 L or .19.

From water dissociation, H+ 5.3 x 10^-14 and
pH= -Log(H+)= 14 - log 5.3 = 13.27 appx