Question

An analytical chemist is titrating 188.5 mL of a 0.9900 M solution of propionic acid (HC,HCO,with a 0.9000 M solution of NaOH. The pK of propionic acid is 4.89. Calculate the pH of the acid solution after the chemist has added 106.3 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=

Answer 1

moles - concen foration (M) X volume (L) milimolen (mmull a concentration (M) X volume (ML) mmol of HC2 H g CO₂ = 0.9900 MX 128.5 mL = 186.615 mmol mmolol NaOH = 0.9000M X 106.3mL = 95.67 mmol Table ICE mmol H2 Hg CO₂ log) + NaOH lag) + NOG H3 CO, lag) + H₂Ole) 1861615 - 95.67 95.67 - 95.67 +95.67 90.945 95.67 Total volume of solution = 1881 5m L. & 10603 mL = 294.8 mL bka of propionic acid - 4.89

90.945 mmol = 0.3085 M In Solution [HC2H5 CO2] = mmol Total volume 294.8 mL [No C2 HS (O2 ) = 95.67 mmol 0.3245M mmol Total volume 294.8 m 2 Using Henderson Howalbalch queron PH= pka & log {Na C 2 H 5 (0₂] = [HC 2 H g CO₂ leko flog / 0,3245 M' 1 0.3085 M) pH = 4.89 + 0.022 = 4.93 pH = 4.93 I Answer Plecue Rate