I am lost on how to calculate all of these. Help is needed, please and thank you

Question

I am lost on how to calculate all of these. Help is needed, please and thank you!!

Please show work.

An experiment was performed to determine the solubility and solubility product constant of lead(II) chloride (gmm=278.10). The equilibrium involved is: PbCl2(s)---> Pb2+(aq) + 2Cl-(aq)

The dissolved chloride in the filtered solution was quantitatively precipitated as AgCl by the addition of excess AgNO3 solution. The solid AgCl was filtered from solution, dried, and weighed. The following data were obtained for three samples of the filtered, saturated solution.

Determination: 1 2 3

Temp of solution, C 20.2 20.4 20.1

volume of PbCl2 solution analyzed, mL 25.00 22.00 20.10

mass of dry AgCl, g 0.2543 0.2276 0.2051

Calculate 1-5 for each determination:

1. number of moles of AgCl(s), Cl-(aq), Pb2+(aq)

2. [Cl-] in PbCl2 solution, M

3. [Pb2+] in PbCl2 solution, M

4. Ksp

5. Solubiity of PbCl2, in g per 100mL

Thank you so much whoever is helping me out! I'm desperate.

Explanation / Answer

(1)# No. of moles of AgCl :

Determination 1 = 0.2543/mol. mass of AgCl

= 0.2543/143.32 = 0.0017 moles

Determination 2 = 0.2276/143.32 = 0.0015moles

Determination 3 = 0.2051/143.32 = 0.0014

# For calculating the moles of [Cl]- :

Ag2+ + Cl-(aq) = AgCl

So, 1 mol of Agcl corrresponds to 1 mole of Cl-(aq) ; Thus same moles of Cl- (aq) will be produced as AgCl

Hence, Determination 1 = 0.0017 moles

Determination 2 = 0.0015moles

Determination 3= 0.0014 moles

#For Pbcl2 , the no. of moles is calculated as follows :

As per the reaction PbCl2 = Pb2+ + 2Cl-

So, 2 moles of Cl- corresponds to 1 mole of Pb2+ (aq)

Hence, Determination 1 = 1/2*0.0017 = 0.00085moles

determination 2 = 1/2*0.0015=0.00075moles

Determination 3= 1/2*0.0014=0.0007moles

(2) M of Cl- calculation :

Determination 1 = 0.0017/0.025 = 0.069 M

Determination 2 = 0.0015/0.022=0.068M

Determination 3= 0.0014/0.02=0.07M

(3) M of Pb2+ calculation : Will be half that of Cl- M in each determinant :

Determination 1= 0.069/2=0.0345M

Determination 2=0.034M

Determination 3=0.035M

(4) Ksp in this reaction = [Pb2+]*[Cl-]^2

Determinant 1 = 0.0345*(0.069)^2= 1.64*10^-4

Determinant 2= 0.034*(0.068)^2=1.57 *10^-4

Determinant 3 = 0.035*(0.07)^2=1.7*10^-4

(5)   solubility=%u221AKsp

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