Question

A 25.0 mL sample of solution of unkown acid, HX, is titrated with 2.87 M KOH. If 59.9 mL of KOH was required to neutralize the sample, find the molaritu of fhe original HX solution.

Answer 1

Number of moles=Molarity x Volume (L)

1000 mL=1 L

We know that when an acid is neutralised completely by a base, the number of moles of acid=number of moles of base

i.e. Molarity of acid x Volume of acid (L)=Molarity of base x Volume of base (L)

Molarity of HX x Volume of HX neutralised (L)=Molarity of KOH x Volume of KOH used (L)

Molarity of HX x 25.0 mL/1000 mL/L=2.87 M x 59.9 mL/1000 mL/L

Molarity of HX=2.87 M x 59.9 mL/25.0 mL=6.88 M

So molarity of HX solution=6.88 M