Chapter 16, Problem 62A

Interpretation Introduction

Interpretation:

The $pH$ of given acid solution is to be calculated for various dilutions.

Concept introduction:

pH gives the concentration of hydrogen ion in a solution.

It tells about acidity and basicity of a solution.

It is given by:

  $\begin{array}{l}pH=-\log \left[H^{+}\right]\\ pOH=-\log \left[O{H}^{-}\right]\\ pH+pOH=14\\ \left[H^{+}\right]isconcentationofhydrogenions\\ \left[O{H}^{-}\right]isconcentationofhydroxylions\end{array}$

Answer to Problem 62A

  $pH$ of $100mL$ 1M $HCl$ is zero.

On increasing the amount of water, hydrogen and hydroxyl ion concentration increases. The increase in hydroxyl ion decreases the acidity of solution.

Explanation of Solution

In strong acid if concentration of solution is greater than ${10}^{-7}M$ , then hydroxyl ions produced by water doesn’t affect much to $pH$ . Thus, $pH$ of $1M$ HCl is given by:

  $\left[H^{+}\right]=1M$

  $\begin{array}{c}pH=-\log \left[H^{+}\right]\\ =-\log 1\\ =0\end{array}$

Thus, $pH$ of $100mL$ 1M $HCl$ is zero.

On addition of water, the volume of solution gets doubled which reduces the concentration to half.

Therefore,

  $\left[H^{+}\right]=0.5M$

  $\begin{array}{c}pH=-\log \left[H^{+}\right]\\ =-\log 0.5\\ =-\log (5\times {10}^{-1})\\ =-\log 5-\log {10}^{-1}\\ =-0.699+1\\ =0.301\end{array}$

Thus, $pH$ of $200mL$

  $HCl$ is 0.301.

Again on doubling the volume, $\left[H^{+}\right]=0.25M$

  $\begin{array}{c}pH=-\log \left[H^{+}\right]\\ =-\log 0.25\\ =-\log (2.5\times {10}^{-1})\\ =-\log 2.5-\log {10}^{-1}\\ =-0.398+1\\ =0.602\end{array}$

Thus, $pH$ of $400mL$

  $HCl$ is 0.602.

The $pH$ will increase on adding water but the value will remain in acidic range of $pH$.

Conclusion

Increase in amount of water in a strong acid decreases the acidic strength.