An analytical chemist is titrating 114.0 mL of a 0.5300M solution of benzoic acid (HC6H5CO2) with a 0.5600M solution of NaOH. The pKa of benzoic acid is 4.20 Calculate the pH of the acid solution after the chemist has added 117.7 mL of the NaOH solution to it. Note for advanced students: you may assume the final volume equals the initial volume of the solution plus the volume of NaOH solution added Round your answer to 2 decimal places.
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First, we need to calculate the moles of benzoic acid and NaOH. Moles of benzoic acid = volume (L) x concentration (M) = 0.114 L x 0.5300 M = 0.06042 mol Moles of NaOH = volume (L) x concentration (M) = 0.1177 L x 0.5600 M = 0.06592 mol Show more…
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