Answer to Question #89106 in General Chemistry for Randall

Question

A student does a titration using KOH and HBr for the purpose of determining the concentration of the HBr. The student places 0.32 M of KOH in the buret and records the initial volume of the base in the buret at 24.3 ml. They then measure out 7.8 ml of acid and place it in a flask under the buret. Indicator is added to the acid. The student then titrates the acid with the base until the acid is neutralized. The final volume of the base is recorded at 33.8 ml. What is the molarity of the acid? The equation for neutralization is given below. Round to two decimal places.

KOH + HBr --> H2O + KBr

Answer 1

Let`s count the volume of used base for titration:

1) 33.8 - 24.3 = 9.5 (ml);

Now we should convert ml to l, because the molar concentration is expressed in mol/l.

9.5 ml = 0.0096 l

Now we calculate the amount of base:

n = 0.32 M * 0.0096 l = 0.003072 mol;

The amount of acid is equal to the amount of base. Now we can calculate the molar concentration (molarity) of acid:

7.8 ml = 0.0078 l;

c = n / V = 0.003072 / 0.0078 = 0.39 M.