Question

Barbituric acid dissociates as follows:

(b)Calculate the pH and fraction of dissociation of 10^-10.00 M barbituric acid.

Short answer

The $\mathrm{pH}\mathrm{of}{10}^{-10.00}\mathrm{M}\mathrm{is}7$

The fraction of disassociation is 99.9%

Step-by-step solution

Step 1 : Definition of pH.

pH is defined as a quantitative measure of acidity and basicity. Compounds having pH more than 7 is acidic in nature while when shown above than 7 is basic in nature. When pH comes 7 , compound is said to be neutral.

Step 2 : Calculate pH and fraction of dissociation.

Since the solution is highly diluted, we can consider that$\mathrm{pH}=7\left(\left[{\mathrm{H}}^{+}\right]={10}^{-7}\right)$.

We will calculate the fraction of dissociation by:

$$\begin{gathered} {\mathrm{K}}_{\mathrm{a}}=\frac{\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{A}}^{-}\right]}{\left[\mathrm{HA}\right]} \\ \left[{\mathrm{A}}^{-}\right]=\frac{{\mathrm{K}}_{\mathrm{a}}}{\left[{\mathrm{H}}^{+}\right]}\times \left[\mathrm{HA}\right] \\ =\frac{9.8\times {10}^{-5}}{{10}^{-7}}\times \left[\mathrm{HA}\right] \\ \left[{\mathrm{A}}^{-}\right]=980\left[\mathrm{HA}\right] \end{gathered}$$

Step 3 : Calculate fraction of dissociation.

Now, using the same reaction, we will find fraction of dissociation:

$$\begin{gathered} \mathrm{a}=\frac{\left[{\mathrm{A}}^{-}\right]}{\left[\mathrm{HA}\right]+\left[{\mathrm{A}}^{-}\right]} \\ =\frac{980\left[\mathrm{HA}\right]}{\left[\mathrm{HA}\right]+980\left[\mathrm{HA}\right]} \\ =980/981 \\ =99.9\% \end{gathered}$$