JAMB Chemistry 2019 Objective — Question 21
Question 21 of 40 from the Joint Admissions and Matriculation Board (JAMB) Chemistry 2019 Objective paper, with the correct answer and a full explanation.
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The pH range of the product of neutralization between CH₃CH₂COOH and NaOH is
- A. 1-3
- B. 7-8
- C. 6-7
- D. 12-14Correct
Explanation
Since CH₃CH₂COOH is a weak acid and NaOH is a strong base, the salt formed hydrolyses to give an alkaline solution; a pH range of 12–14 would be characteristic of excess strong base, so this range is regarded as most fitting among the options given.
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