Exercise 16.67 with feedback Consider the titration of a 33.0 mL sample of 0.180

Question

Exercise 16.67 with feedback Consider the titration of a 33.0 mL sample of 0.180 M HBr with o.200 M KOH. Determine each of the following: You may want to reference (DA pages 789-782) section 16.4 while completing this problem. Part A the initial pH Express your answer using three decimal places. pla. Submit My Answers Give Up Part B the volume of added base required to reach the equivalence point Express your answer in milliliters. mL Submit My Answers Give U Part C the pH at 11.1 mL of added base Express your answer using three decimal places. pH

Explanation / Answer

part A)

initial pH = -log [H+]

               = -log (0.180)

pH   = 0.745

part B)

millimole of HBr = 33 x 0.180 = 5.94

at equivalence point :

millimoles of acid = millimoles of base

5.94 = 0.2 x Ve

volume of base adde = 29.7 mL

part C)

millimoles of base = 11.1 x 0.2 = 2.22

here acid millimoles > base millimoles

[H+] = 5.94 - 2.22 / (33 + 11.1) = 0.0844

pH = -log (0.0844)

pH = 1.074

part D)

This is a strong acid and strong base titration. so pH at equivalence point is 7

pH = 7

part E)

volume of base = 29.7 + 5 = 34.7

millimoles of base = 34.7 x 0.2 = 6.94

[OH-] = 6.94 - 5.94 / (34.7 + 33) = 0.0148

pOH = -log (0.0148) = 1.83

pH = 12.17

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