Chapter U6, Problem 19STP

Interpretation Introduction

Interpretation:

The pH value of a 1.0 M solution of alizarin yellow Hln needs to be determined.

Concept introduction:

The pH of a weak acid can be determined using the ICE table. This is used to calculate the equilibrium concentration of all the species present in the solution.

The pH of the solution can be calculated from equilibrium concentration of hydrogen ion.

$\text{pH}=-\log \left[{\text{H}}^{+}\right]$.

Answer to Problem 19STP

The pH is 5.2 thus, option (B) is correct.

Explanation of Solution

The given ionization of alizarin yellow is as follows:

$\text{Hln}\left(aq\right)\rightleftharpoons {\text{H}}^{+}\left(aq\right)+{\text{In}}^{-}\left(aq\right)K=3.4\times {10}^{-11}$

The given molarity or concentration of solution is 1.0 M. The ICE table can be represented as follows:

$\begin{array}{l}\text{ Hln}\left(aq\right)\rightleftharpoons {\text{H}}^{+}\left(aq\right)+{\text{In}}^{-}\left(aq\right)\\ I\text{ 1}\text{.0 - -}\\ C-x\text{ +}x\text{ +}x\\ E1.0-xxx\end{array}$

The equilibrium expression can be written as follows:

$K=\frac{\left[{\text{In}}^{-}\right]\left[{\text{H}}^{+}\right]}{\left[\text{HIn}\right]}$

Putting the values,

$3.4\times {10}^{-11}=\frac{\left(x\right)\left(x\right)}{\left(1-x\right)}$

Since, the value of equilibrium constant is very low, the value of x can be neglected from denominator.

$3.4\times {10}^{-11}=\left(x\right)\left(x\right)$

Or,

$x=\sqrt{3.4\times {10}^{-11}}=5.83\times {10}^{-6}$

Thus, the concentration of hydrogen ion in the solution is $5.83\times {10}^{-6}$.

Now, the pH of the solution can be calculated as follows:

$\text{pH}=-\log \left[{\text{H}}^{+}\right]$

Putting the values,

$\text{pH}=-\log \left(5.83\times {10}^{-6}\right)=5.2$

Conclusion

Thus, the pH of the solution is 5.2 and correct option is (B).