Problem: Calculate the pH of a solution when 45.0mL of 0.100M NaOH (a strong base) is added to 50.0mL of 0.100M CH3COOH (a weak acid).
HINTS:
We found that the number of mols of NaOH is 4.50 x 10^-3 and the number of mols of CH3COOH is 5.00 x 10^-3.
Since each mol of NaOH causes the weak acid to dissociate, the number of mols of CH3COO- is equal to 4.50 x 10^-3 and the remaining number of mols of CH3COOH is 5.00 x 10^-4.
We can find the Molarity of CH3COO- and CH3COOH by dividing the moles by the total volume, 0.095L.
We discussed in class that as the concentration of the conjugate base, CH3COO-, increases, it will react with water and release OH- ions. This results in the solution becoming more basic. Please note that the Ka accounts for this!! So, we simply use the equation for Ka to solve for [H+].
