Question

Jackie diluted 7.5 mL weak acid with 7.5 mL distilled water. Then she titrated the diluted weak acid with 0.100 M NaOH. It takes 22.52 mL of NaOH to reach the equivalence point. What is the original concentration of the weak acid? What is the diluted concentration of the weak acid?

Answer 1

1st calculate the diluted concentration of weak acid.

Balanced chemical equation is:

NaOH + HA ---> NaA + H2O

Here:

M(NaOH)=0.1 M

V(NaOH)=22.52 mL

V(HA)=15.0 mL

According to balanced reaction:

1*number of mol of NaOH =1*number of mol of HA

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

1*0.1*22.52 = 1*M(HA)*15.0

M(HA) = 0.150 M

This is concentration of diluted acid

Now find the concentration of original acid

use dilution formula

M1*V1 = M2*V2

1---> is for stock solution

2---> is for diluted solution

Given:

M2 = 0.15 M

V1 = 7.5 mL

V2 = 15 mL

use:

M1*V1 = M2*V2

M1 = (M2 * V2) / V1

M1 = (0.15*15)/7.5

M1 = 0.3 M

original concentration of the weak acid = 0.30 M

diluted concentration of the weak acid = 0.15 M