Question

O ACIDS AND BASES Calculating the pH of a weak acid titrated with a strong base Try Again Your answer is wrong. In addition to checking your math, check that you used the right data and DID NOT round any intermediate calculations. An analytical chemist is titrating 2010 ml of a 1.100 M solution of benzoic acid (HCH,CO) with a 0.6800 M solution of NaOH. The pK, of benzoic acid is 4.20. Calculate the pH of the acid solution after the chemist has added 147.6 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 - 12.54 5 ?

Answer 1

Required reaction :

HC₆H₅CO₂ + NaOH ----------------> NaC₆H₅CO₂ + H₂O

No.of milli moles of HC₆H₅CO₂= (201 mL) x (1.1 M) = 221.1 m.moles

No.of milli miles of NaOH = (147.6 mL) x (0.68 M) = 100.368 m.moles

Hence, No.of moles of HC₆H₅CO₂ left after reaction = (221.1-100.368) m.moles = 120.732 m.moles

No.of moles of salt formed = 100.368 m.moles

Total volume = 201 mL + 147.6 mL = 348.6 mL

[salt] = (100.368 m.moles) / (348.6 mL) = 0.2879 M

[acid] = (120.732 m.moles) / (348.6 mL) = 0.346 M

Therefore, According to Henderson Hasselbalch equation, pH is given by:

pH = pKa + log [salt]/[acid]

pH = (4.20) + log (0.2879) / (0.346)

pH = (4.20) + (-0.080)

pH = 4.1 ----------------- (ANSWER)