Question

10.00mL of a sample of concentrated hydrochloric acid is diluted to 100.00mL. If a 20.00 mL portion of this diluted acid required 15.23mL of 1.5 M NaOH to reach the endpoint, what is the concentration of the original, undiluted HCl?

Answer 1

1st find the concentration of HCl in diluted solution.

Balanced chemical equation is:

NaOH + HCl ---> NaCl + H2O

Here:

M(NaOH)=1.5 M

V(NaOH)=15.23 mL

V(HCl)=20.0 mL

According to balanced reaction:

1*number of mol of NaOH =1*number of mol of HCl

1*M(NaOH)*V(NaOH) =1*M(HCl)*V(HCl)

1*1.5*15.23 = 1*M(HCl)*20.0

M(HCl) = 1.1422 M

Now use dilution formula to find concentration in stock solution.

use dilution formula

M1*V1 = M2*V2

1---> is for stock solution

2---> is for diluted solution

Given:

M2 = 1.1422 M

V1 = 10 mL

V2 = 100 mL

use:

M1*V1 = M2*V2

M1 = (M2 * V2) / V1

M1 = (1.1422*100)/10

M1 = 11.422 M

Answer: 11.4 M