35.0 mL of a 0.250 M solution of KOH is titrated with 0.150 M HCl. After 35.0 mL of the HCl has been added, the resultant solution is:

Question

A) Acidic and after the equivalence point
B) Basic and after the equivalence point
C) Neutral and at the equivalence point
D) Basic and before the equivalence point
E) Acidic and before the equivalence point

Answer 1

millimols HCl = M x L = 0.150 x 35 = 5.25
millimols KOH = M x L = 0.250 x 35 = 8.75
........HCl + KOH ==> KCl + H2O
.......5.25.....8.75
It should be obvious that all of the KOH has not been neutralized by the HCl so the solution will be basic. It will take another 8.75-5.25 = ? mols HCl to finish the neutralization

Answer 2

So, it should be letter D, basic and before the equivalence point??