Absorbance: A) 0.8286 B)0.5025 C)0.5359 Equation: A=2524.5[CrO4]+ 0.0096 Solutio
ID: 705757 • Letter: A
Question
Absorbance: A) 0.8286 B)0.5025 C)0.5359
Equation: A=2524.5[CrO4]+ 0.0096 Solutions: A) 0.10M NaNO3 B)4.00x10^-4M AgNO3/ 0.10M NaNO3 C) 3.00x10^-5 M K2CrO4/ 0.10M NaNO3
-Use the equilibrium [CrO4] and the intial common ion concentrations it compute an ICE chart
-use the values from the E line to compute Ksp for Ag2CrO4 A) 0.10M NaNO3 B)4.00x10^-4M AgNO3/ 0.10M NaNO3 C) 3.00x10^-5 M K2CrO4/ 0.10M NaNO3
-Use the equilibrium [CrO4] and the intial common ion concentrations it compute an ICE chart
-use the values from the E line to compute Ksp for Ag2CrO4
-calculate the average value of Ksp and the standard deviation Please shows completed ICE charts A = 2524, 5 CrO2+ g09 K,, Determination Three solutions are stirring with ex?ess solid silver chromate on the rear bench of the laboratory. We may assume these solutions are saturated at equilibrium with the excess silver chromate. Solution A: 0.10M NaNO3 Solution B: 4.00 x 10- M AgNOs/ 0.10M NaNO3 Solution C: 3.00x 10- M K2CrO4 /0.10M NaNO3 Each group should measure the absorbance of each solution on your spectrophotometer. Record the results.
Explanation / Answer
Given: Callibration equation :A=2524.5 [CrO42-] +0.0096
1) Data for absorbance values:
solution 1) A=0.8286
[CrO42-]=(A-0.0096)/2524.5=(0.8286-0.0096)/2524.5=3.244*10^-4M
solution 2)A=0.5025
[CrO42-]=(A-0.0096)/2524.5=(0.5025-0.0096)/2524.5=1.952*10^-4M
solution 3) A=0.5359
[CrO42-]=(A-0.0096)/2524.5=(0.5359-0.0096)/2524.5=2.085*10^-4M
2) ICE table
solution A) AgCrO4(s)+0.1M NaNO3, no common ion effect
Solution B) AgCrO4(s)+0.1M NaNO3+AgNO3
AgNO3(s) <--->Ag+(aq) +NO3- (aq) ,Ksp=6.0*10^-4=[Ag+][NO3-]
But [NO3-]=0.10M
6.0*10^-4=[Ag+][NO3-]=[Ag+](0.1M)
[Ag+]=(6.0*10^-4)/0.1=6.0*10^-3 M (maximum solubility )
[Ag+]initial=6.0*10^-3 M+2x (from Ag2CrO4 dissociation)
Solution C)AgCrO4(s)+0.1M NaNO3+K2CrO4
K2CrO4(s)-->2K+(aq)+CrO42-(aq) [complete dissociation]
[CrO42-]initial=3.00*10^-5M+x (from AgCrO4 dissolution)
Ag2CrO4<-->2Ag+(aq) +CrO42-(aq)
Solution 1)
x=[CrO42-]=3.244*10^-4M
[Ag+]=2[CrO42-]=(2*3.244*10^-4M)=6.488*10^-4M
Ksp(Ag2CrO4)=[Ag+]^2[CrO42-]
Ksp(Ag2CrO4)=[Ag+]^2[CrO42-]=(6.488*10^-4)^2(3.244*10^-4)=1.365*10^-10
Solution 3)
x=[CrO42-]=2.085*10^-4M
[Ag+]=2[CrO42-]=(2*2.085*10^-4M)=4.170*10^-4M
Ksp(Ag2CrO4)=[Ag+]^2[CrO42-]=(4.170*10^-4)^2(2.085*10^-4)=3.625*10^-11
Solution 2)
x=[CrO42-]=1.952*10^-4M
[Ag+]=2[CrO42-]=(2*1.952*10^-4M)=3.904*10^-4M
Ksp(Ag2CrO4)=[Ag+]^2[CrO42-]=(3.904*10^-4)^2(1.952*10^-4)=2.975*10^-11
4) Average KSp=1/3((1.365*10^-10)+(3.625*10^-11)+(2.975*10^-11))=6.75*10^-11
[AgCrO4](s) [Ag+] [CrO42-] Initial NA 6.0*10^-3 M 3.00*10^-3M change NA +2x +x Equilibrium(x) NA 6.0*10^-5 M+2x 3.00*10^-5M+x Eqm