Question

Calculate the pH of a solution prepared by dissolving 0.170 mol of benzoic acid and 0.310...

Calculate the pH of a solution prepared by dissolving 0.170 mol of benzoic acid and 0.310 mol of sodium benzoate in water sufficient to yield 1.50 L of solution. The Ka of benzoic acid is 6.30 × 10-5. Please show steps to help me understand this.

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Answer #1

first caluculate the molarity of benzoic acid and molarity of benzoate using the formula

Molarity M = no of moles / volume in liters

Molarity of benzoic acid = 0.170 / 1.5 = 0.1133 M

molarity of sodium benzoate = 0.310 / 1.5 = 0.2067 M

now calulate pKa value from the Ka

pKa = -log[Ka]   = -log[6.30 × 10-5]   = 4.2

now substitute all these values in the below equation

direct equation

pH = pKa + log [sodium benzoate/ benzoic acid]

pH = 4 + log [0.2067/0.1133]

pH = 4 + 0.261

pH = 4.261

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