Question

10 Tuesday, March 31 x Chem162 HW7b - Solubility equilibria - Graded and timed- Homework 2 3 4 5 Post Question 5 of 5 Previous Next 537. mL of 0.00238 M Nal (aq) is combined with 498. mL of 0.00467 M Pb(NO3)2 (aq). Determine if a precipitate will form given that the Ksp of Pbly is 1.40x10-8 O Precipitation will occur because Osp Kep O O Precipitation will occur because Osp >ksp Precipitation will not occur because Qsp > Kap Precipitation will not occur because Qsp Kap Precipitation will occur because Qsp - Ksp O

Answer 1

we have to deduce the concentration of Pb and I in new solution

after we add the both given solution there will be incresse in volume so for the Ksp of PbI2 we need concentration of Pb and I in new obtained solution by mixing two given solutions

2nd option is correct

see image below

Date : - Far (Na I) 537 mi 0.00238 M 537+ 498 me me we know formula mivimo vo 537 X 0.00 238 - M x 1035 [m2 = 0.001234 New I concentration conec For Pb(NO3)₂ = 498 me, yo = 837 +498 m-7 m, = 0.00467 Miya M2 V 2.0 0.00467 x 498 = M2 x 1035. My = 0.002247 New Pt Po I2 - Pot + 2 I Ksp [P6+2] [I'd Here we have. Qsp = 2.772 x 10-6 Qsp 7 KSP Precipitation occurs.

Answer 2

PAGE NO DATE [Na] = 0.00238M [ Po(wo;)] = 0.00467 M V NI = 537 ml - Upolno) = 498 ul. Now, according to statement of question, NaI es combined with Pblowood. le we need to get new concentrations of both Na I and Pb(NO) Using = (M. v) initial = (M. v) food. [NaII fine = 0.00238 X537 [Pb(NO₃),] = 0.00467X 498 (537 +498) [NaI] = 0,00123 M [Pb(NO3)2) = 0.00224 M final (537 +498) - final Now, Na I + Pb(NO₂), 2 Pbig + 2 alano, Y = 0.00123 M het = 0.00224 strog electrolyte) T Pb I. (s) Pb2+ (ap) + 2 I (op) - = [Pb 2+ ] [ Ij2 Not at equilibrium Op = (0,00124) (0,00123) 0.0024 M 0,00123 M dse = 3.39 x 10-9 Rsp = 1.40 x 10-8 dsp < Ksp => Precipitation will not occur,