Question

1) Chloropropionic acid, ClCH₂CH₂COOH is a weak monoprotic acid with K_a = 7.94 x 10^-5 M. Calculate the pH at the equivalence point in a titration 40.4 mL of 0.49 M chloropropionic acid with 0.9 M KOH.

2) What is the pH after 22.90 mL of NaOH are added?

Answer 1

At equivalence point Noº of m-moles of chlorop ropionic acid = No of mo-moles of kom Given Concentrat id = c = 0.49M Volume of acid = v. = 40.4mL No of m.mole of acid = cv = 0.49 MX 40.4mL 19.79m-mele Given Concentration of koH = = o-gM Suppose volume of Koh be V, ml cv =ąva 19.79 momole = V = 31.98 mb 0.9M X V mL Total volume of solution = (V, TV2) = ( 40.4 + 21-98) mL V = 62.38mL किब No of m.mole of sult foomed - No' of om moles of acid reacted - 5 19.79 m.mole CICH CH COOH (aq) at Kohlag) - CICH, cry cookt t House Sult | concen toution of salt foomed = 19.79 m.mole 0-317244 62.38 mL CICH CH cookt is salt of weak acid and strong Base. It will undergo hydrolysis in presence of water CICHCHcookt 1094 + you cloud con la q + Hotaq pka = -logka = -log (7.94x105) pka = 4.10

PH = ½ pkw + pka + loge] -_=[14 + 9-10 + logo-3132) [PH = 8.80 ) Answa Addition of 22.90mL KOH No of monoles of koH = 0.9M* 22-90mL = 20.61m-mole No of m.mole of and = 19.79m-mole - > No. of m. mole of kom left = (20.61-19-49) m-mole = 0.82m-mol Total Volume = (22.90 + 40.4) mL = 63.3mL Concentration of kom = 0.82 m-mole 63-30t = [kor [-OM] = 0.01295 M 0.01295M pOH pOH = -log[-oh = -log (0.01295M) = 1-88 = 14 - 1-88 = 12-11 Answer PH PH