2. A 0.5520 g of potassium hydrogen phthalate, KHCH&O4, a monoprotic weak acid, takes 24.90 mL of NaOH solution to reach the equivalence point. a. What is the molarity of the NaOH? b. Then, the base was used to analyze an unknown acid. It required 30.20 ml of the base to reach the equivalence point using 1.100 grams of the unknown. Calculate the equivalent weight of the unknown acid. C. If the unknown is prepared as KHP plus impurity, calculate the % of KHP in the unknown.
2. A 0.5520 g of potassium hydrogen phthalate, KHCH&O4, a monoprotic weak acid, takes 24.90 mL of NaOH solution to reach the equivalence point. a. What is the molarity of the NaOH? b. Then, the base was used to analyze an unknown acid. It required 30.20 ml of the base to reach the equivalence point using 1.100 grams of the unknown. Calculate the equivalent weight of the unknown acid. C. If the unknown is prepared as KHP plus impurity, calculate the % of KHP in the unknown.
Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
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![2. A 0.5520 g of potassium hydrogen phthalate, KHC&H&O4, a monoprotic weak acid, takes 24.90 ml of
NaOH solution to reach the equivalence point.
a. What is the molarity of the NaOH?
b. Then, the base was used to analyze an unknown acid. It required 30.20 ml of the base to reach
the equivalence point using 1.100 grams of the unknown. Calculate the equivalent weight of
the unknown acid.
c. If the unknown is prepared as KHP plus impurity, calculate the % of KHP in the unknown.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F714ec5cc-ca4e-4c5b-814c-de61801b113c%2F6ef37765-2590-4d1e-b4e4-581e918c9bde%2Fa2tlnhq_processed.jpeg&w=3840&q=75)
Transcribed Image Text:2. A 0.5520 g of potassium hydrogen phthalate, KHC&H&O4, a monoprotic weak acid, takes 24.90 ml of
NaOH solution to reach the equivalence point.
a. What is the molarity of the NaOH?
b. Then, the base was used to analyze an unknown acid. It required 30.20 ml of the base to reach
the equivalence point using 1.100 grams of the unknown. Calculate the equivalent weight of
the unknown acid.
c. If the unknown is prepared as KHP plus impurity, calculate the % of KHP in the unknown.
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