Question

A 14.50 mL sample of nitric acid (HNO₃) is titrated to the end point by the addition of 10.45 mL of a 1.525 M solution of barium hydroxide (Ba(OH)₂). What is the molarity of the nitric acid solution? (Balanced equation: 2HNO3 + Ba(OH)2 = Ba(NO3)2 + 2 H20)

Answer 1

Balanced chemical equation is:

Ba(OH)2 + 2 HNO3 ---> Ba(NO3)2 + 2 H2O

Here:

M(Ba(OH)2)=1.525 M

V(Ba(OH)2)=10.45 mL

V(HNO3)=14.5 mL

According to balanced reaction:

2*number of mol of Ba(OH)2 =1*number of mol of HNO3

2*M(Ba(OH)2)*V(Ba(OH)2) =1*M(HNO3)*V(HNO3)

2*1.525*10.45 = 1*M(HNO3)*14.5

M(HNO3) = 2.1981 M

Answer: 2.198 M