Question

1.In a titration, 50.0 mL of an acetic acid solution required 20.0 mL of a standard solution of
0.200 mol/L NaOH(aq). The concentration of the acetic acid solution is
a.
0.08 mol/L.
b.
0.50 mol/L.
c.
0.05 mol/L.
d.
1.0 mol/L.
e.
0.30 mol/L.
How do I solve this question?

Answer 1

You do these titration questions one of two ways.
IF you know (and only if) that 1 mol acid = 1 mol base (as is true in this case), you may us mLacid x Macid = mLbase x M base. Substitue and solve for the one unnown.

If you don't know you have a 1:1 ratio, you MUST do it another way, as follows for this problem. You may want to try it both ways to see that you get the same answer both ways.
1. Write the equation. HAc is acetic acid . HAc + NaOH ==> NaAc + H2O

2. mols NaOH - M x L = ?

3. Using the coefficients in the balanced equation, convert mols NaOH to mols HAc. In this case mols HAc = mols NaOH. IF you had a titration of say H2SO4 and NaOH, then
H2SO4 + 2NaOH ==> Na2SO4 + 2H2O.
Here you would bet mols NaOH the same way (mols = M x L) but to convert to mols H2SO4, then mols H2SO4 = 1/2 mols NaOH.

4. Then, M HAc = mols HAc/L HAc = ?

Post your work if you get stuck.