Question

The osmotic pressure of $0.1$ M monobasic acid solution is $0.11$ $RT$ at '$T$ $K$'.

What is it's $pH$ value?

Answer 1

The expression for the osmotic pressure, $\Pi$ is $\Pi =iCRT$. Here, $i$ is the Van't Hoff factor, C is the concentration of the acid, R is the ideal gas constant and T is the temperature in K.

But the osmotic pressure is $0.11RT$. Hence, $\Pi =iCRT=0.11RT$

or $iC=0.11$ or $i=\frac{0.11}{C}=\frac{0.11}{0.1}=1.1$. Thus, the Van't Hoff factor is 1.1.

The expression for the degree of dissociation, $\alpha$ is $\alpha =\frac{i-1}{n^{\prime }-1}$. Here, $n^{\prime }$ is the number of ions produced by complete dissociation of 1 molecule of monobasic acid. It is 2. Substituting the values in the above expression, $\alpha =\frac{1.1-1}{2-1}=0.1$

The hydrogen ion concentration is $\left[H^{+}\right]=\alpha \times C=0.1\times 0.1=0.01$

The pH of the solution is $pH=-log\left[H^{+}\right]=-log\left(0.01\right)=2$.