Question

An analytical chemist is titrating 238.8 mL of a 1.200 M solution of benzoic acid (HCH CO2) with a 1.000 M solution of NaOH. The pk, of benzoic acid is 4.20. Calculate the pH of the acid solution after the chemist has added 333.5 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. pH = 0 = xo? X 6

Answer 1

238.8 ml of 1.2 m benzoic acid = 238.8 m2x12 mom aniola 238.8 x 1.2 mol of benoic 1000 m li - 0.28 656 mol of benzoic and 333.5 mlot im NaOH = 333.5 mLxim of muolt 333-5% i mog of much Todo not = 0.9335 mol of Nuolt 0.3335 mol of mooth react with 0.3335 mol ph cook to form 0-3335 mol of colts coona sout Honce the soln contain (0.3335-0.28656) mot. or, doubly mol of exam Noot. istot - - mol of muott. volume of mor total sole 18 (238-8 +333s)me of www. Class 0.04694 moldin olt Bolt 572.3x10 -3 [ of Nuus o 8. 2x10-8 m of moth SHE

[OH-] = 8.2x10-2 M Here the solution contains excess of was. Hence the pH of the solution is more or less equal to that excess of or Naot only. Neglecting small contribution coming from hydrolysis. polt - - Log tot] = -log (8.2x102) = 1.09 pH = 14 - por = 14-1.09 = 12.91 in pH of the solution is pro 12.91