Calculate the pH of a buffer solution prepared by dissolving 23.4 g benzoic acid

Question

Calculate the pH of a buffer solution prepared by dissolving 23.4 g benzoic acid (HC7H5O2) and 35.0 g sodium benzoate in 215.0 mL of solution.

Explanation / Answer

Calculate the pH of a buffer solution prepared by dissolving 23.4 g benzoic acid (HC7H5O2) and 35.0 g sodium benzoate in 215.0 mL of solution. ANSWER: C6H5COOH < ------------> C6H5COO- + H+ Step 1: Convert Mass to moles, Moles of Benzoic acid = 23.4 g/122.12g/mol = 0.192 moles of Benzoic acid Moles of Sodium benzoate = 35.0g/144.1 g/mol = 0.243 moles of Sodium benzoate. Step 2: Calculate Molarity (Volume is 215.0 ml or 0.215 L). Molarity of Benzoic acid = 0.192 moles/0.215 L =0.893M Molarity of Sodium benzoate = 0.243 moles/0.215 L =1.13M Step 3: Calculate pKa. pKa = -log (Ka) of Benzoic acid (Ka = 6.5 x10^-5). = -log (6.5 x10^-5) = 4.19 Step 4: Calculate pH. pH = pKa + log[C6H5COO-]/[C6H5COOH] pH = 4.19 + log (1.13M/0.893M) = 4.19 + 0.102 = 4.292 The pH of this buffer solution is 4.292.

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