Question

What volume of 0.075 M Ba(OH)2(aq) is required to completely neutralize 22.1 mL of 1.48 M HNO3(aq)?

A) 1.7 x 10–3 L B) 0.87 L C) 1.2 x 10–3 L D) 0.22 L E) 4.9 x 10–3 L

Answer 1

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

1mole         2 mole

Ba(OH)2                                                          

M1 = 0.075M

V1 =

n1 = 1 mole

HNO3

M2 = 1.48M
V2 = 22.1ml

n2 = 2mole

for neutralisation ,        M1V1/n1 = M2V2/n2

                               V1 = M2V2n1/n2M1

                               V1 = 1.48 x 22.1 x 1/2x0.075

                               V1 = 218.05mL

                               V1 =0.218 L = 0.22L

The answer is D.