Chapter 22, Problem 21STP

To determine

The concentration of $\text{HCl}$ with respect to that of $\text{NaOH}$ .

Answer to Problem 21STP

The $\text{HCl}$ solution is twice as concentrated as the $\text{NaOH}$ solution.

Explanation of Solution

Given:

Volume of $\text{HCl}$ = $25.0\text{ ml}$

Volume of $\text{NaOH}$ required to neutralize $25.0\text{ ml}$ of $\text{HCl}$ = $50.0\text{ ml}$

Calculation:

As per the problem,

Let the concentration of $\text{HCl}$ be $x\text{ mole/ml}$ and that of $\text{NaOH}$ be $y\text{ mole/ml}$ .

  $\text{HCl}$ is a strong acid and $\text{NaOH}$ is a strong base. They both dissociate completely to produce ${\text{H}}^{+}$ and ${\text{OH}}^{-}$ ions.

For neutralization, the concentration of ${\text{H}}^{+}$ and ${\text{OH}}^{-}$ ions should be equal in the solution, that is, at neutralization, $1{\text{mole/ml of H}}^{+}\text{}\text{ =}\text{}1{\text{mole/ml of OH}}^{-}$

Therefore,

  $\begin{array}{l}25\text{ ml of }x\text{ mole/ml HCl= 50 ml of }y\text{ mole/ml NaOH}\\ \text{25 ml}\cdot x\text{ mole/ml}=50\text{ ml}\cdot y\text{ mole/ml}\\ x\text{ mole/ml}=\frac{(50\text{ ml)(}y\text{ mole/ml)}}{25\text{ ml}}\\ x\text{ mole/ml}=2y\text{ mole/ml}\end{array}$

Conclusion: The $\text{HCl}$ solution is twice as concentrated as the $\text{NaOH}$ solution.