A solution has [HC₇H₅O₂] = 0.100 M and [Ca(C₇H₅O₂)₂] = 0.200 M. Ka = 6.3 × 10^-5 for HC₇H₅O₂. The solution volume is 5.00 L. What is the pH of the solution after 10.00 ml of 5.00 M NaOH is added? A) 4.80 B) 4.86 C) 4.65 D) 4.70 E) 4.75

Question

Quizplus · Accepted Answer

First, we need to write an equation for the dissociation of HC₇H₅O₂: HC₇H₅O₂ ⇌ H+ + C₇H₅O₂ The equilibrium constant (Ka) is given as 6.3 × 10^-5. Next, we need to calculate the initial concentration of H+. Since the solution is 0.1 M in HC₇H₅O₂, we can assume that initially, all of it is in the form of HC₇H₅O₂. Therefore, [H+] = [HC₇H₅O₂] = 0.1 M. Adding NaOH will react with H+ to form water and Na+. The balanced equation is: NaOH + H+ ⇌ Na+ + H2O We can assume the initial concentration of NaOH and Na+ is negligible compared to the solution volume. After 10.00 ml of 5.00 M NaOH is added, the final volume of the solution is 5.010 L (since 10.00 ml is negligible compared to 5.00 L). The amount of moles of NaOH that was added to the solution is: (10.00 mL) * (5.00 mol/L) * (1 L/1000 mL) = 0.0500 mol Since the reaction between NaOH and H+ is a 1:1 stoichiometry, the amount of moles of H+ that was neutralized is also 0.0500 mol. Therefore, the new concentration of H+ is: [H+] = (moles of H+ originally) - (moles of H+ neutralized by NaOH) / (final solution volume) [H+] = (0.1 M) - (0.0500 mol / 5.010 L) [H+] = 0.09 M Now we can use the equilibrium constant (Ka) and the initial concentrations of H+ and HC₇H₅O₂ to calculate the pH: Ka = [H+][C₇H₅O₂] / [HC₇H₅O₂] 0.000063 = (0.09 M)(x) / (0.1 M - x) Solving for x, we get: x = 1.99 × 10^-6 pH = -log[H+] = 4.70 (rounded to two decimal places)

A Solution Has [HC7H5O2] = 0

A Solution Has [HC₇H₅O₂] = 0