Question

Using activity coefficients, find the pHand fraction of dissociation of 50.0 mM LiBr Take the size of ${\mathbf{C}}_{\mathbf{6}}{\mathbf{H}}_{\mathbf{5}}{\mathbf{O}}^{\mathbf{-}}$ to be 600 pm.

Short answer

The pH of 50.0 mM hydroxybenzene (phenol) in 0.050 M LiBr is 5.64.

The fraction of dissociation of 50.0 mM hydroxybenzene (phenol) in 0.050 M LiBr is 0.0053%

Step-by-step solution

Step : Definition of pH

pH is defined as a quantitative measure of acidity and basicity.The solutions having less than pH are acidic, and the solutions with pH more than 7 are basic, and the neutral solutions have a pH of 7.

Step : Calculation of x value

$$\begin{gathered} K_a=\frac{\left[H^{+}\right]y\left(H^{+}\right)\left[A^{-}\right]Y\left(A^{-}\right)}{\left[HA\right]y\left(HA\right)} \\ 1.01\times {10}^{-10}=\frac{\left(x\right)\left(0.86\right)\left(x\right)\left(0.835\right)}{\left(0.05-x\right)\left(1.00\right)} \\ x=2.65\times {10}^{-6} \end{gathered}$$

Step : Calculating  and fraction of dissociation

$$\begin{gathered} \mathrm{pH}=-\log \left[{\mathrm{H}}^{+}\right]\mathrm{y}\left({\mathrm{H}}^{+}\right) \\ =-\log \left(\left(2.65\times {10}^{-6}\right)\left(0.86\right)\right) \\ =5.64 \end{gathered}$$

$$\begin{gathered} \alpha =\frac{\left[A^{-}\right]}{\left[HA\right]+\left[A^{-}\right]} \\ =\frac{2.65\times {10}^{-6}}{0.05} \\ =5.3\times {10}^{-5} \\ \alpha =0.0053\% \end{gathered}$$