Please help with parts d and e ASAP!!! Thank you and will rate back!!Determine the affect of spectator ions on the pH of a solution of a weak acid.a. What is the pH of a 0.0200 M benzoic acid (Ka = 6.28 x 10-5) solution in water?b. What is the % dissociation of the benzoic acid in pure water?c. If the benzoic acid is dissolved in 0.10 M CaCl2 instead of pure water: Calculate the ionic strength of the solution (assume that the contribution from the benzoic acid is negligible).d. Now set up the equilibrium expression with the activities of the ions. Calculate the concentration of both the H+ and the benzoate ion. What is the pH of this solution? What is the % dissociation?e. By what % did the benzoic acid dissociation change between the pure water solution and the calcium chloride solution? **Educational Content: Calculation of Hydrogen Ion Concentration in a Weak Acid Solution**In this exercise, we explore the dissociation of benzoic acid (PhCOOH) in an aqueous solution and calculate the concentration of hydrogen ions, \([H^+]\), using the known acid dissociation constant, \(Ka\).**Given Data:**- Concentration of benzoic acid, \(C = 0.02 \, M\)- Acid dissociation constant, \(Ka = 6.28 \times 10^{-5}\)**Chemical Equation:**The dissociation of benzoic acid is represented as follows:\[\text{PhCOOH} \rightleftharpoons \text{PhCOO}^- + \text{H}^+\]**Concentration Changes:**- At \(t = 0\), initial concentrations are: - \([\text{PhCOOH}] = C\) - \([\text{PhCOO}^-] = 0\) - \([\text{H}^+] = 0\)- At equilibrium, concentrations become: - \([\text{PhCOOH}] = C - C\alpha\) - \([\text{PhCOO}^-] = C\alpha\) - \([\text{H}^+] = C\alpha\)**Expression for \(Ka\):**\[ Ka = \frac{C\alpha \cdot C\alpha}{C(1-\alpha)} \]This simplifies to:\[ Ka = \frac{C\alpha^2}{1 - \alpha} \]Since \(\alpha \ll 1\), we approximate \(1 - \alpha \approx 1\), thus:\[ Ka = C\alpha^2 \]**Calculation of Degree of Ionization (\(\alpha\)):**\[\alpha = \sqrt{\frac{Ka}{C}} = \sqrt{\frac{6.28 \times 10^{-5}}{0.02}} = 0.056\]**Calculation of \([H^+]\):**\[[H^+] = C\alpha = 0.02 \times 0.056 = 1.12 \times 10^{-3} \, M\]Thus, the hydrogen ion concentration in the solution is \(1.12 \times ### Calculation of pH and Ionic Strength#### (a) Calculating pHGiven the hydrogen ion concentration:\[\text{pH} = -\log[H^+]\]Substitute the value:\[\text{pH} = -\log(1.12 \times 10^{-3})\]Therefore:\[\text{pH} = 2.95\]#### (b) Percent Dissociation of Benzoic Acid in Pure WaterThe percent dissociation is given by:\[\alpha \times 100\%\]Where \(\alpha = 0.056\):\[= 0.056 \times 100\%\]Thus, the percent dissociation is:\[= 5.6\%\]#### (c) Ionic Strength CalculationThe ionic strength, considering the contribution from benzoic acid as negligible, is determined primarily from the 0.1 M \( \text{CaCl}_2 \) solution.The formula for ionic strength \( I \) is:\[I = \frac{1}{2} \sum C_i z_i^2\]For \( \text{CaCl}_2 \):- \( \text{CaCl}_2 \) dissociates into \( \text{Ca}^{2+} \) and \( 2 \text{Cl}^- \).- Initial concentration of \( \text{CaCl}_2 \) is 0.10 M, leading to concentrations: \( \text{Ca}^{2+} = 0.10 \) M and \( \text{Cl}^- = 0.20 \) M.The ionic strength is calculated as follows:\[I_{\text{CaCl}_2} = \frac{1}{2} \{ (0.10 \times 2^2) + (0.20 \times 1^2) \}\]\[= \frac{1}{2} \{ 0.40 + 0.20 \}\]Thus:\[= 0.30 \, \text{M}\]This detailed breakdown illustrates the computation of pH, percent dissociation, and ionic strength for educational purposes.

We have to calculate the following ..... a) the pH of a 0.0200 M benzoic acid (Ka = 6.28 x 10-5) solution in water. b) the % dissociation of the benzoic acid in pure water. c) we have to calculate Calculate the ionic strength of the solution (assume that the contribution from the benzoic acid is negligible) If the benzoic acid is dissolved in 0.10 M CaCl2 instead of pure water.