Question

1.A total of 15.0 mL of 0.10 M NaOH is added to 25.0 mL of 0.10 M HCl in a flask during titration. What is the new pH of the solution?
A. 1.0
B. 1.6
C. 12.4
D. 13.0

2. Imagine an experiment in which 0.10 M HCl is added 1.0 mL at a time to a conical flask that contains 25.0 mL of 0.10 M KOH solution. Calculate the pH of the solution after 5.0 mL of 0.10 HCl has been added.
A.1.2
B.3.3
C.10.7
D.12.8

Answer 1

For #1, we first need to figure out how many moles of NaOH and HCl we have in order to determine what has reacted and what we have left. For NaOH there are 0.015 L x 0.10 M = 0.0015 mol and for HCl there are 0.025 L x 0.10 M = 0.0025 mol.

NaOH and HCl react in a 1:1 ratio according to this equation:
NaOH + HCl --> NaCl + H2O

This means the 0.0015 mol of NaOH neutralizes the same number of moles of HCl. This leaves us with 0.0025 - 0.0015 = 0.0010 mol of HCl. The volume of the solution is now 0.015 + 0.025 = 0.04 L, so our new concentration of HCl is 0.0010 mol / 0.04 L = 0.025 M. The pH of the solution is therefore -log(0.025) = 1.6 and your answer is B.

For #2 you can do a very similar thing - first find moles of your reactants, then see what's used up by the reaction and figure out how much is left, then calculate your new [H+} and pH! Just follow the steps I showed for the first one!

Hope that helps!