Calculating the pH of a weak acid titrated with a strong b- An analytical chemis

Question

Calculating the pH of a weak acid titrated with a strong b- An analytical chemist is titrating 87.0 mL of a 0.4200 M solution of butanoic acid (HC,H,CO2) with a 0.8800 M solution of KOH. The p K, of butanoic acid is 4.82. Calculate the pH of the acid solution after the chemist has added 44.0 mL of the KOH solution to it. Note for advanced students: you may assume the final volume equals the initial volume of the solution plus the volume of KOH solution added. Round your answer to 2 decimal places.

Explanation / Answer

no of mol of butanoic acid = 87*0.42/1000 = 0.0365 mol

no of mol of KOH added = 44*0.88/1000 = 0.03872 mol

no of mol of KOH excess = 0.03872- 0.0365 = 0.00222 mol

Concentration of excess KOH = n/v

           = 0.00222/(87+44)*1000

           = 0.017 M

pH = 14 - (-log(OH-))

     = 14 - (-log(0.017))

     = 12.23

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