Question

Be sure to answer all parts. A 35.00−mL solution of 0.2500 M HF is titrated with a standardized 0.1561 M solution of NaOH at 25 ° C. (a) What is the pH of the HF solution before titrant is added? (b) How many milliliters of titrant are required to reach the equivalence point? mL (c) What is the pH at 0.50 mL before the equivalence point? (d) What is the pH at the equivalence point? (e) What is the pH at 0.50 mL after the equivalence point?

Answer 1

- The Reaction between HF HF (aq) + Nach lag Before Addition of titront and Naoh is shown below Naf (aq) + Holl Concentration of HF = C, = 0.2500M Volume of HE = V = 35.com ka = Acid diseociations constant of HE 26-6X164 HFlaq) + Houll - Hotlaq + Frage Initial oom OM - 0.2500(1-) 0-2500X 0 .2500X 0. 2 OM AH Equilibrium [Hot FT - 10.25002) [HFJ 0-2500 (1-2) ka= 0.2500 1-dal[v. weak a cid] = 0.25ood? x= 6-68164 0-25 5.14x10² = D.oslu. - = 0.01285 [Hot] =0.25004 = 0,2500 XD.oslum PH = -log[hot] = -log(0-01285M) PH = 1.89] Answer

16 Given Ho of m mole of HF = 0.2500M X 35mL. e 8-75m-male | Given Concentration of NaoH = 0.1561M Volume of NaOH = V mL At Equivalence point No of m.moles of HF = Not of m.mole of NaOH. 8.75m.mol = 0.156M X Y mL L ₂ = 56.08 mL Answer 10 0.50m before equivalence point Volume of NaOH required = ( 56.08-0-50 mL = 55.58ml No of m-mole of NaOH = 0.1561 M X 55-58mL - 8-676momole No of m-mole of H F = 8.75m-mole. No of m-mole of He left = (8.45-8-676) m-mole = 0.074 m.nole - Total volume of solution - (35 + 55-58 ) mL = 90.58ml pka = -logka = -log (6.6x104) = 3.18 & No of monoles of salt (NAF) formed = 8.676m-mole PH = pka + logh Naf LHE = + 8-6760amale 30.58 ml 3.18 tlog 0.074m-mole/ samut IPH = 5.24 Answer

(d) At equivalence point whole acid is neutralized by base & No of m.mole of salt formed = How of momole of acid reacted 1 - 8-75 m-mok - Total volume = v= (35+ 56.08)mL = 91.08mL Concentration of salt formed = 8.75mmole 91.08mL = 0.09606M = c. The salt formed will hydrolyze in presence of water Naflags + HOW) = Hilac) + OH(aq) PH = Pkw + pka +logc] [co Hydrolysis of salt of weak acid and Strong Base 14 + 3.18 tlogo.ogrom 960M PH = 8.08) Angue