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11. What equation or expression would be used to calculate the pH of a solution prepared by dissolving 0.280 mol of benzoic acid (C7H5O2H) and 0.120 mol of sodium benzoate (NaC7H5O2) in water sufficient to yield 1.00 L of solution? The Ka of benzoic acid is 6.50 x 10-5. A) Ka expression в) кЬ еxpression c) Kw = Ka*Kb D) Henderson Hasselbach Equation Calculate the pH of the solution above.

11. What equation or expression would be used to calculate the pH of a solution prepared by dissolving 0.280 mol of benzoic acid (C7H5O2H) and 0.120 mol of sodium benzoate (NaC7H5O2) in water sufficient to yield 1.00 L of solution? The Ka of benzoic acid is 6.50 x 10-5. A) Ka expression в) кЬ еxpression c) Kw = Ka*Kb D) Henderson Hasselbach Equation Calculate the pH of the solution above.

Chemistry
Chemistry
10th Edition
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
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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**Question 11:** What equation or expression would be used to calculate the pH of a solution prepared by dissolving 0.280 mol of benzoic acid (C₇H₅O₂H) and 0.120 mol of sodium benzoate (NaC₇H₅O₂) in water sufficient to yield 1.00 L of solution? The \( K_a \) of benzoic acid is \( 6.50 \times 10^{-5} \). A) \( K_a \) expression B) \( K_b \) expression C) \( K_w = K_a \times K_b \) D) Henderson Hasselbach Equation Calculate the pH of the solution above.
Transcribed Image Text:**Question 11:** What equation or expression would be used to calculate the pH of a solution prepared by dissolving 0.280 mol of benzoic acid (C₇H₅O₂H) and 0.120 mol of sodium benzoate (NaC₇H₅O₂) in water sufficient to yield 1.00 L of solution? The \( K_a \) of benzoic acid is \( 6.50 \times 10^{-5} \). A) \( K_a \) expression B) \( K_b \) expression C) \( K_w = K_a \times K_b \) D) Henderson Hasselbach Equation Calculate the pH of the solution above.
Expert Solution
Step 1

Given : moles of weak acid i.e benzoic acid = 0.280 mol

And moles of conjugate base salt of weak acid i.e sodium benzoate = 0.120 mol

Since the solution has both weak acid and its conjugate base salt.

Hence the solution is a buffer solution.

 

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