Question

An analytical chemist is titrating 228.6 mL of a 0.3500 M solution of nitrous acid (HNO.) with a 1.000 M solution of NaOH. The pk, of nitrous acid is 3.35. Calculate the pH of the acid solution after the chemist has added 59.90 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= pH= xs ?

Answer 1

No.of milli moles of acid = 228.6 mL * 0.3500 M = 80.01 m.moles

No.of milli moles of base = (59.90 mL) * (1 M) = 59.90 m.moles

Hence, Concentration of salt formed = (59.90 m.moles) / (228.6+59.90 ) = 0.208 M

Concentration of acid = (80.01 - 59.90) / (228.6+59.90 ) = 0.0697 M

Therefore, pH of the solution is given by: Henderson Hasselbalch equation:

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

pH = 3.35 + log (0.208 M) / (0.0697 M)

pH = 3.35 + (0.474)

pH = 3.8 -----------(ANSWER)