Ahhh, last chemistry question for homework...?

Question

What is the pH of a solution formed by mixing 125.0mL of 0.0250M HCl with 75.0mL of 0.0500M NaOH?

AHHH please explain so I can understand in the future... :)

Answer 1

125.0 mL x 0.0250 M = 3.125 millimoles (0.003125 moles) NaOH

75.0 mL x 0.0500 M = 3.750 millimoles (0.003750 moles) HCl

NaOH + HCl --> NaCl + H2O

Excess HCl = 0.00375 - 0.003125 = 0.000625 mole HCl

Total solution volume = 125 + 75 = 200 mL

Molarity of excess HCl = 0.000625 mol / 0.200 L = 0.003125 M

pH = -log(0.003125) = 2.51