Homework Answers
38) The answer will be 0.075 M (Option A).
Explanation: First equivalence point occurs when 15 mL of NaOH is added as there is a sharp increase in the pH. At this point, carboxyl group of the amino acid is fully neutralized. We assume that the stoichiometric ratio of NaOH & glycine at this point is 1:1. Now, we can easily calculate molarity of glycine by using M1V1 = M2V2.
According to the question, volume of glycine (V1) = 20 mL, molarity of NaOH (M2) = 0.1 M, volume of NaOH (V2) = 15 mL
So, molarity of glycine (M1) = M2V2 x 1/V1 = 0.1 x 15 x 1/20 = 0.075 M
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Question 38 If the strong base solution used in the titration curve shown in Figure 2...
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