Question

Exactly 50.0mL of HOCl solution of unknown concentration was titrated with 0.100 mol NaOH. An end point was reached when 38.5mL of the base was added. Calculate the molar concentration of the HOCl soluion.

Answer 1

Answer:

Molarity of HOCl =0.87 mol/Litre

Explanation:

It is a acid base titration after completion of reaction salt and water.

At the end point of raction,

mili equivalent of HOCl = Mili equivalent of NaOH;

$normality = molarity\times n-factor$

$N_1\times V_1=N_2\times V_2$

n-factor for HOCL and NaOH is 1 for both

hence

normality = molarity;

molarity of $NaOH=\frac{mole}{volume of solution in litre}$

$molarity =\frac{0.1Mol}{0.0885L} = Normality=N_2$

$N_1 \times50=\frac{0.1}{0.0885} \times38.5$

$N_1=0.87$;

Molarity of HOCl =0.87 mol/Litre