A. Preparation of a Calibration Curve. Initial [Cron .0024 Volume of 0.0024 M K

Question

A. Preparation of a Calibration Curve. Initial [Cron .0024 Volume of 0.0024 M K CrosTotal Volume CrO1 in solution Absorbance 1. ImL 2. 2mL 3. 3mL 4.4mL 5. 5mL 100mL 100mL 100mL 100mL 100mL 2.4x10-s 4.8x10-S 7.2x10-5 9.6x10- 1.2x10 0.02 0.04 0.06 0.09 0.14 asorbance 0.16 0.34 0.12 0.1 0.08 0.06 0.04 0.02 y 1208.3 LADE-04 B. Determination of the Solubility-Product Constant Write the KsR expression of Ag CrO List the theoretical value of the Ksp of Ag.CrO [As]2[CrO4] Chart Area 1.12x10-12 Ag'T 7.68x10-+ Absorbance ICro1 3.84x10 Kse of Ag:CrO 0.32 2x3.84x 10-4 | [7.68x-4]2[3.84x-4] 2.26x10-10 0.33 3.96x102x3.96x10-2113 2.48x10-10 Average Kan (show calculations) Calculate the percentage error in you experimentally determined value for K

Explanation / Answer

from part 1)

Average Ksp=(Ksp1+Ksp2)/2=((2.26*10^-10)+(2.48*10^-10))/2=2.37*10^-10

Percentage error=(difference between experimental and theoretical value)/theoretical value   *100

=((2.37*10^-10)-(1.12*10^-12))/(1.12*10^-12) *100=210.607*10^2 %

Post lab

Q1) if [Ag+]=0.0024M (limiting reactant)

[CrO42-]=0.004M(excess reactant)

Ag2CrO4 (s) <--->2Ag+ (aq) +CrO42-(aq)

then [Ag+] at equilibrium <<<2[CrO42-] that will decrease the amount of CrO42- at equilibrium. thus, Ksp will be much lesser.

Ksp =[Ag+]^2 [CrO42-]

Q2) If cuvette is dirty ,then Absorbance will be less so will be the concentration of Chromate ion that is proportional to the value of Absorbance.

Abs=el*[CrO42-]

So [Ag+]=2[CrO42-] will also be lesser

Ksp will again have a small value

Q3)Ksp=2.37*10^-10=[Ag+]^2 [CrO42-]

Let [CrO42-]=S,[Ag+]=2S

2.37*10^-10=(2S)^2*S=4S^3

S=3.898*10^-4 mol/L

V=10ml=0.010L

Solubility=0.010L*(3.898*10^-4 mol/L)=3.898*10^-4 mol

Mass of Ag2CrO4=3.898*10^-4 mol*molar mass=(3.898*10^-4 mol)*331.73g//mol=0.129g=129mg

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