Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

No Service 3:25 PM LON-CAPA Pre-Final Assign-Q7 capa2.its.dal.ca A buffer soluti

ID: 694617 • Letter: N

Question

No Service 3:25 PM LON-CAPA Pre-Final Assign-Q7 capa2.its.dal.ca A buffer solution is prepared by mixing 44.6 mL of 0.0597 M sodium hydrogen citrate with 55.1 mL of 0.380 M sodium citrate A table of pKa values can be found here 1. Calculate the pH (to two decimal places) of this solution Assume the 5% approximation is valid and that the volumes are additive 6.82 Submit Answer Incorrect. Tries 2/3 Previous Tries 2. Calculate the pH (to two decimal places) of the buffer solution after the addition of 215 mL of a 0.00335 M solution of barium hydroxide to the existing buffer solution. Assume the 5% approximation is valid and that the volumes are additive 6.02 Submit Answer Incorrect. Tries 1/3 Previous Tries his discussion is closed Send Feedback

Explanation / Answer


1) pH of buffer = pka2 + log(sodium citrate / sodium hydrogen citrate)

   pka2 of citric acid = 4.761

no of mol of sodium citrate taken = 55.1*0.38 = 20.938 mmol

no of mol of sodium hydrogen citrate taken = 44.6*0.0597 = 2.663 mmol

pH = 4.761+log(20.938/2.663)

    = 5.65

2) no of mol of Ba(OH)2 added = 215*0.00335 = 0.72 mmol

   pH of buffer = pka2 + log(sodium citrate+2*Ba(OH)2 / sodium hydrogen citrate -2*Ba(OH)2)

pH = 4.761+log((20.938+2*0.72)/(2.663-2*0.72))

     = 6.0234

1) molarity of mixture(M) = M1V1+M2V2/V1+V2

             = ((5.18*10^-5)*829+(1.17*10^-4)*77.5)/(829+77.5)

             = 5.74*10^-5 M

pH = -log[OH-]

   = -log(5.74*10^-5)

   = 4.24

Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at
drjack9650@gmail.com
Chat Now And Get Quote