posted by Bert on .
If 100mL of a 0.10M NaOH solution is added to 75mL of a 0.15 HCl solution, then what is the pH of the resultant solution?
Write the equation.
HCl + NaOH ==> NaCl + H2O
mols NaOH = M*L = 0.10 M x 0.1 L = ??
mols HCl = M*L = 0.15 M x 0.075 L = ??
See which is in excess, how much is left over after reacting, then pH = - log(H^+). Post your work if you get stuck.
I don't understand where the H+ comes from, I mean I know it comes from the acid, but don't I need an equilibrium constant K to solve for the H+?
No. Since the product is NaCl and water and the NaCl doesn't hydrolyze, the H+ or OH- comes from the excess of HCl OR NaOH. Look at the mols HCl and the mols NaOH. Calculate each according to my last post. You will see one is larger than the other. So subtract them to find the difference and that will be the H+ if it is the HCl in excess OR it will be the OH- if it is the NaOH in excess. Both HCl AND NaOH are strong acids and bases so neither has a Ka or Kb.
A note that when you subtract the difference is the MOLS of excess reagent. The concentration is mols/L to calculate H+ or OH- so mols/0.175 L is the concentration.