ej 12/1/2017 12:55 AM 62.1/1001 1/22/201702:18 AM Gradebook O Assignment Informa
ID: 582918 • Letter: E
Question
ej 12/1/2017 12:55 AM 62.1/1001 1/22/201702:18 AM Gradebook O Assignment Information eh Print Calculator Periodic Table Question 39 of 39 Available From: 10/4/2017 11:55 P Map do Sapling Learning Due Date: 12/1/2017 12:55 AM 1.602 g of N-2-acetamido-2-amingethane-suifonic acid potassium salt (ACES K', MW 22029gmol) is dissolved in 87.27 mL of water. 17.69 mL of HCI is added to the solution, resulting in a pH of 6.51. Calculate the concentration of the HCI solution. The pka of ACES is 6.85 Points Possible: 100 Grade Category Graded Description: Policies Number Homeworlk HCI] = You can check your answers. You can view solutions when you comple give up on any question. You can keep trying to answer each quest until you get it right or give up You lose 5% of the points available to each answer in your question for each incorrect attempt at that answer eTextbook O Help With This Topic O Web Help & Videos Previous Give Up & View Solution O Technical Support and Bug Reports Check Answer Next Exit HintExplanation / Answer
pH = pKa + log [ACESK]/[ACESH]
6.51 = 6.85 + log [ACESK]/[ACESH]
-0.34 = log [ACESK]/[ACESH]
0.457 = [ACESK]/[ACESH]
Number of mmoles of added = [mass (g)/molar mass(g)] x 1000 = 1.602 x 1000/220.29 = 7.272 mmol
Suppose, concentration of HCl = C mole/liter
Number of mmoles of HCl added = M x V (mL) = 17.69 C
Number of mmoles of ACESH formed = Number of mmoles of HCl added = 17.69 C
Number of mmoles of ACESK remaining = 7.272 - 17.69 C
0.457 = [ACESK]/[ACESH]
0.457 = (7.272 - 17.69 C) /17.69 C
8.08433 C = 7.272 - 17.69 C
25.77433 C = 7.272
C = 0.282 M
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.