Question

1. 200 mL of an aqueous solution contains 0.030 M concentrations of both Pb2+ and Ag+. If 100

mL of 6.0 x 10-2 M NaCl is added to this solution will a precipitate form? If so, what will the

precipitate be? The Ksp values for PbCl2 and AgCl are 1.7 x 10-5 and 1.8 x 10-10] 2. Which of the following is the expression for the solubility product of Ba3(AsO4)2? 3. The pKa of a weak acid is 6.50. What is the value of Ka for this acid?

1. 200 mL of an aqueous solution contains 0.030 M concentrations of both Pb2+ and Agt. If 100 mL of 6.0 x 102 M NaCl is added to this solution will a precipitate form? If so, what will the precipitate be? The Ksp values for PbCl2 and AgCl are 1.7 x 10-5 and 1.8 x 10-10) 2. Which of the following is the expression for the solubility product of Baz(AsO4)2? a. Ksp=[Ba²*1*/AsO2?)? b. Ksp=[3 x Ba2+]*[2 x AsO43-12 c. Ksp=3[Ba2+] * 2 AsOA?) d. Ksp=3[Ba2+] + 2(AsOA?? e. Ksp= (Ba?*)s[AsO4%)2 3. The pk, of a weak acid is 6.50. What is the value of Ka for this acid?

If all 10 questions could be answered that would be great!

Answer 1

a 15 2t 0-03 x 0.2 = 0.006 Volume of solution initially = 200ml = 0.21 concentration of Pb²+ and Ag t = 0.63M No. of moles of Pb No of moles of Ay 100 ml of 6x10² M Na co į added. No of moles of ce = 6x 10 On 006 + -0.006 20 -2 6x 10 xo, Final concentrations 25 27 O. 3 0.006 = 0.02 M O 006 3 0.02M 0.3 0.006 0.02M 0.3 Pb Na lagt Ice like consider the reactions, Pbce. Z Pb Agce. ² Ag+ + ce 30 27 * 2ce

Pbce has Qap [Palce (0.02) (0.02) 8x 10-6 0 Poce, has Kop = 17x10-5 it will not 1 Since Qs < Kap for Pbce precipitate Now; for Agce, Psp = (Ag + ſce- 10 0,02x0.02 . & 4x1004 For Agce, hip = 1.8x10*** & Since, Pop Kap for Age, it will form précipitate? .: A precipitate will form and that precipitate is Agll. 20 (As out 2+ 3- 2 3 Ba + 2As 2-4 Ba Kap 25 As Of 013-12 is pka 30 -6.50 Ka = 10-PK 10 = 3.162x10-7

Camlin Page Date 4) 500 ml = Molarity 0-5 Molecular wt of acetic acid = 60 Molecular wit of sodium acetate = 82 Volume of solution = 0.51 No. of moles of acetic acid= 15 0.25 No of moles of sodium acetate = 3.25 = 3.96x10 Molarity of acetic acid = 0.25 = 8.5M of sodium acetate 3.96X10 7.92x10-2 M Ka of acetic acid = 108x105 na pka - log (1.8x10-5) = 4.74 the Henderson- Hasselbach equation, [A] & pH = pke + log THAT where [A] = concentration of sodium acetate 5.7. 92x 10^? THA = concentration of acetic acid 7.42x102 pH= 15 - log ka Now we use 201 o.5 25 4.74 + loge = 3.94

Date 1 s) 2.5M solution of NH Ce fag) K of NH₃ = 1.8x10-5 We know o Ka XR assuming standard 70-74 * conditions. te 10 + ions by Ki far INH, +) = 5.56x10 1.8x10-5 NH, Cl dissolves in solution to form NHL which act as weak acid. protonating water to form ammonia, and hydronium ions Hot (aq) (ол) +И, Оe) 4 lag) + Hot (as) Ka 2,0*x (NH) NH₃ (ag) NHET NH (aq) (NH A 20 z 25 can 5:56 x 10 fit, 04*(NH) (2.5) 3 1.398109 Hot] assume that one molecule of hydronium must form for every of ammonia chemical equation. Also Ka is small, so x <<2) that forms, according to the [Hot] = V1.398104 = 3.728x10-5 A pH = -log (3,728x10*4)= 4:43 30

Page Date - Cestain acid pka =4.88 101 -4.88 = 1.318x10 Acetic acid pka = 4.74 10 10-4. 1.74 1.8192x10-5 Ka of certain acid < Ke of acetic acid is the certain acid is weaker than acetic acid. 7 20 - log [Ht log (4.6710-8 7.3 pot = -6.7 [H]= 4.6x10-8 pH = 14 - PH = 6.7 [OH-] = 10 31.99510 M the sample is basic (slightly) since pH=73. 8) | Bacolle] 10 = 2 2 [(OH)3 = 2x0 . 095 0.1 M pok 12 Ha. 14-pOH = 14.-) = 13 Answer =o.o5M - log Toti - log 0.1 = e

Camlin Page Date 1 а - weak acid Э K, х (се, соон ) (0) [и] acidic buffer од си, соон /си, соо®N, (from Chycoona) си, соои : Сн,соог эн* К = |-8х705 К. = си, соо] (н [си, соор (и си, соо] - log Kat log (са,соод) [Сң соо] 7 - log Ka - log (си, соби) [CH₂ coo: Henderson (pka + log (CH coo Hasselbach (Сн, соор squation Va (weak acid] (conjugate base] — (си, сеон | [ch coo pH pka + log! PK+= ek Н. Spka 4.74 ) nh,t +H, 0 NH, OH+H* (Radarwegian С. 6х10 s. 6xlo _ [ми, 1 since [NH, OH] = [NHA conjugate acid Ә = pH |-- Я - Коу (1-gro-) B ) би зми, о . S. Axis"" weak base

Carlin Page Date 1 fuceo (H+) = 5-6x 10ml pH = log (5 6 x 10.-0.742-9-252 s [ceo pH = pka = -log (3.2x10) = 7.495 D (INO,] = [NO, ) pH = pka = -log (4.5+10%) = 3.346 El Nace / He tont be It will not be buffer because because Hee acid & thus it appreciable change in pH on addition of acid os base. answee in fasta (Chicoona/ CHCOOH) since its pH is 4.74 which is within the required range of 4.6-4.8 sterong, will give 201 so 10 false Stronger acids have smaller pka values than weak acids.