24.00 mL of a 0.25 M NaOH solution is titrated with 0.10M HCl. What is the pH of the solution after 24.00 mL of the HCl has been added?

Answer 1

millimoles NaOH = mL x M = 24.00 mL x 0.25 M = 6.00
millimoles HCl = 24.00 mL x 0.10 M = 2.40
total volume = 48.00 mL

.................................NaOH + HCl ==>NaCl + H2O
initial.........................6.00.........0............0.........0
added.....................................2.40............................
change.................... -2.40.....-2.40.........+2.40....+2.40
equilibrium.................3.60.........0..............2.40.......2.40
The NaCl contributes nothing to the pH of the final solution. The pH is determined by the excess of NaOH present. (NaOH) = millimoles/mL = 3.60/48.00 = ? M = (OH^-)
pOH = -log (OH^-). Then
pH + pOH = pKw = 14.00. You know pOH from above and you know pKw = 14; substitute and solve for pH. Post your work if you get stuck.