2. (a) Write the equations needed to find the pH of and the concentrations of th
ID: 1034225 • Letter: 2
Question
2. (a) Write the equations needed to find the pH of and the concentrations of the major species in a solution of 0.0500M potassium benzoate. Include activity coefficients where appropriate. (b) Reduce the equations you wrote in part (a) to a single equation with the (H O*] as the only unknown quantity. You may use "F as the formal concentration of the solution and you may neglect activities in this part of the calculation. Do not make any further simplifying assumptions. (Rearrange you final equation so that all of your terms are equal to zero.) (c) Enter your final equation (and appropriate constant values) in Excel and use Excel Goal Seek (What If? Analysis) to solve for [H:0]. Before using Excel Goal Seek, click the Microsoft Office Button, then Excel Options, and select Formulas. In Calculations Options, set Maximum Change to le-14. Also make sure to enable iterative calculations.) Once you solve for [Hs0*], use that value to find the concentrations of all other species in solution and the pH. You should submit all of your work (not just the Excel work) for this problem showing all the steps of the systematic treatment of this system. Also, show your work from Excel-print the formulas that you used to solve for your final answers (don't just print the spreadsheet that shows the final answers.)Explanation / Answer
potassium benzoate [BzOK] = 0.05 M
BzO- + H2O <==> BzOH + OH-
let x amount hydrolyzed
Kb = [BzOH][OH-]/[BzO-]
1 x 10^-14/6.3 x 10^-5 = x^2/0.05
x = [OH-] = [BzOH] = 2.82 x 10^-6 M
[H3O+] = 1 x 10^-14/2.82 x 10^-6 = 3.55 x 10^-9 M
pH = -log[H3O+]
= -log(3.55 x 10^-9)
= 8.45
with activity coefficient
ionic strength (u) = 1/2(0.05 x 1^2 + 0.05 x 1^2)
= 0.05
[H3O+] = inv.log[(-0.51 x 1^1 x sq.rt.(0.05)/(1 + 3.3 x 0.9 x sq.rt.(0.05)]
= 0.854
[OH-] = inv.log[(-0.51 x 1^1 x sq.rt.(0.05)/(1 + 3.3 x 0.35 x sq.rt.(0.05)]
= 0.812
Kb = [BzOH][OH-]/[BzO-]
1 x 10^-14/6.3 x 10^-5 = x^2(0.812)^2/0.05
x = [OH-] = [BzOH] = 3.47 x 10^-6 M
[H3O+] = (1 x 10^-14 x 0.854)/3.47 x 10^-6 = 2.46 x 10^-9 M
pH = -log[H3O+]
= -log(2.46 x 10^-9)
= 8.61
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.