Question

Titration of 0.7865 g sample of pure potassium hydrogen phthalate requires 35.73 ml of NaOH solution. Calculate the molarity of the NaOH solution. This acid base reaction is:

NaOH+KHC8H4O4->NaKC8H4O4+H2O

Answer 1

m(KHPh) = 0.7865 g;

M(KHPh) = 204 g/mol;

n(KHPh) = m(KHPh)/M(KHPh) = 0.7865/204 = 0.0039 mol;

V(NaOH) = 35.73 mL = 0.03575 L;

M(NaOH) = 40 g/mol;

NaOH + KHPh = NaKPh+ H₂O;

By the chemical reaction:

n(NaOH) = n(KHPh) = 0.0039 mol;

C(NaOH) = n(NaOH)/M(NaOH) = 0.0039/0.03575 = 0.11 M.