# Finding the concentration of hydrochloric acid by titration

In a class experiment, $\pu{20 mL}$ of hydrochloric acid was diluted with distilled water to $\pu{500 mL}$. $\pu{20 mL}$ of this diluted solution was added to $\pu{20 mL}$ of $\pu{0.1 M}$ sodium hydroxide in a conical flask. Potassium permanganate was added into the flask until it turned a light pink colour. $\pu{15.6 mL}$ was needed.

The information I know so far is: $$\ce{NaOH + HCl -> NaCl + H2O}$$ \begin{align} c &= \frac{n}{V}\\ n &= c \cdot V\\ c(\ce{NaOH}) &= \pu{0.1 M}\\ V(\ce{NaOH}) &= \pu{0.02 L}\\ n(\ce{NaOH}) &= \pu{0.1 M}\times \pu{0.02 L} \end{align}

I am not understanding how it is possible to work out the concentration of hydrochloric acid from the conducted experiment and the known information, and I am confused as to how the volume of potassium permanganate allows this concentration to be found.

• What is the strength of $KMnO_4$ solution ??? – Soumik Das Feb 13 '18 at 8:23

For,calculating strength of the hydrochloric acid by titration, you must have to know the strength of $\ce{KMnO4}$.
Let's assume the strength of $\ce{KMnO4}$ was $x\ \mathrm M$.
The $\ce{HCl}$ was titrated with $\ce{NaOH}$ of $0.1\ \mathrm N$.After titrating, if the solution decolurises $\ce{KMnO4}$, it means that there are excess $\ce{HCl}$ present which is now acting as a reducing agent as per the equation, $$\ce{2MnO4- + 10Cl- + 16H+ <=> 5Cl2 + 2Mn^2+ + 8H2O}$$ So, in $20\ \mathrm{ml}$ of $500\ \mathrm{ml}$ dilution, moles of $\ce{HCl}$ present = $(0.1 \cdot 20 + 15.6 \cdot x \cdot 5) \cdot 10^{-3}=y$ (say.)
So in total $500\ \mathrm{ml}$, moles of $\ce{HCl}$ present = $y \cdot 500/20=z$ (say.)
The actual $\ce{HCl}$ was in $20\ \mathrm{ml}$. So, strength of the actual $\ce{HCl}$ solution =$z \cdot 1000/20\ \mathrm M$ (or $\mathrm N$).
For example if $x=0.1$, strength of $\ce{HCl}$ will be = $12.25\ \mathrm M$.