Question

You have titrated 25.0 mL of acid A with 65.35 mL of a 0.231 molar solution of base B. The mole ratio is 1 A = 1 B. What is the concentration of the solution of A?

The answer is 0.604,but  i don't know the process

Answer 1

a. no of moles = molarity * volume in liters

for NaOH => 0.10 M * 0.013 L => 0.0013 moles

b. acid and the base are in 1:1 ratio means

no of moles are same so the moles of acid is 0.0013 moles

c. using the dilution law

M1V1 = M2V2

for NaOH => M1 = 0.10 M , V1 = 13 mL

for Acid => M2 = ? , V2 = 25 mL

M2 = M1V1 / V2

M2 = 0.10 * 13 / 25

M2 = 0.052 M

molarity of acid is 0.052 M