NaOH is a strong base; it reacts with CH3COOH, a weak acid, as follows:
CH3COOH + NaOH ==> CH3COONa + HOH
Since you began with 10 millimoles CH3COOH and 5 millimoles of NaOH, you will produce 5 millimols of the salt (CH3COONa) and you will have left over, at equilibrium, 5 millimoles CH3COOH un-reacted.
You have a solution of a weak base and its salt. So why is this a buffer?
Acetic acid ionizes like this:
CH3COOH + H2O ==> CH3COO^- + H3O^+
Sodium acetate ionizes like this:
CH3COONa ==> CH3COO^- + Na^+
Here's the skinny.
If you try to change the pH of the solution by adding a strong base (NaOH for example), the OH^- are neutralized by the free acetic acid in solution and the pH changes very little. If you try to change the pH by adding a strong acid (HCl for example), the acetate ions (from the sodium acetate) react as follows:
CH3COO^- + H^+ ==> CH3COOH to form acetic acid which just adds a little more acetic acid to what whatr there already. The pH changes little. Within reason, then, addition of strong acid or strong base changes the pH very little. That's why it's a buffered solution.
A buffered solution consists of a weak acid and its salt OR a weak base and its salt. An example of the first isl acetic acid and sodium acetate; an example of the second is ammonia and ammonium chloride.
The last part of your question is yes if you make it. That is, notice the problem started out with acetic acid (an excess) and a strong base (NaOH) which MAKES the salt and leaves some of the acetic acid around. The only other kind I can think of is ammonium acetate. This is the salt of a WEAK acid (acetic acid) AND a WEAK base (NH3)(aq).
thank you....i understand the concept clearly now.