I am not really sure how to do this:
Calculate the pH when 1.0*10^-3 moles of solid NaOH is added to
a) One liter of 9.1*10^ -6 M HCL
b) One liter of a solution containig 0.05 M CH3COOH and 0.1 M CH3CooNa
For a), calculate moles HCl (Moles = M x L) and see which is in excess. Since both HCl and NaOH are strong electrolytes, their molarity will be the H^+ (or OH^- if NaOH is in excess) and pH can be determined from that.
For b), it is more complicated than that.
Acetic acid and sodium acetate constitute a buffered solution. The Henderson-Hasselbalch equation prevails. So find moles of the acetate and moles acetic acid, then add the amount of NaOH. NaOH, of course, reacts with the acetic acid as follows:
CH3COOH + NaOH ==> H2O + CH3COONa
So the CH3COOH is decreased by the amount of NaOH added and the CH3COONa is increased by the amount of NaOH added. Plug those new numbers into the H-H equation to find the new pH. The purpose of this problem is for you to compare how much adding a strong base, NaOH, to an un-buffered solution (a) and to a buffered solution (b). The change is large to a) but not to b).