Post Utme Rsu 2019 Chemistry Question 2

Practice objective / multiple choice question 2 from the 2019 Post Utme Rsu Chemistry examination.

Post Utme Rsu 2019 Chemistry Objective / Multiple Choice
Question 2 POST_UTME_RSU • 2019 • CHEMISTRY • objective

A 25.0 mL sample of 0.100 M HCl is titrated with 0.100 M NaOH. Calculate the pH of the solution after 25.0 mL of NaOH has been added.

Answer Options

Correct Answer A. \( pH = \frac{1}{2} \log \frac{[H^+] [OH^-]}{[H_2O]} \)
B. \( pH = \frac{1}{2} \log \frac{[H_2O]}{[H^+] [OH^-]} \)
C. \( pH = \frac{1}{2} \log \frac{[H^+] [OH^-]}{[H_2O]} \)
D. \( pH = \frac{1}{2} \log \frac{[H_2O]}{[H^+] [OH^-]} \)
Correct Answer
A
Correct Option:
\( pH = \frac{1}{2} \log \frac{[H^+] [OH^-]}{[H_2O]} \)

About This Question

This is Post Utme Rsu 2019 Chemistry Question 2. It is one of the objective questions from the 2019 Post Utme Rsu Chemistry examination.

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