Molar solubility and Ksp??? Hello Everyone Ok so I did a Kidney Stone Lab in Che

Question

Molar solubility and Ksp???

Hello Everyone Ok so I did a Kidney Stone Lab in Chem 2 in which we took different moles of HCL and 1 trial of lemon juice and dissolved a rock (Calcium Phosphate) inside it. The rock is .1g of Calcium phosphate. I created Calcium phosphate by combining 5.24g of Na3PO4 and 5.32g of CaCl2. I mixed it with 100ml of water. My actual yeild was 2.891g.

I used to .1M HCL (20ml), 1M HCL(5ml) and 3M HCL(1ml) to dissolve the rock (3 seperate trials). As well as lemon juice (7ml) to dissolve the rock (Another trial).

I do not know how to find the ksp or the molar solubility of the calcium phosphate stone in HCL.

Please show your work and the answer. I need this to solve other problems.

Thank you so much.

Explanation / Answer

Ca3(PO4)2 =====> 3Ca+2 + 2PO4-3

Ksp for Ca2(PO4)3 will be Ksp= [Ca+2]^3[PO4^-3]^2

Now we need to find out the concentration

Hence, Molarity = weight of solute in g x 1000 / ( molecular weight of solute x volume of solvent in mL

For first trial, total volume =100 + 20 mL = 120 mL

Molar mass of Ca2(PO4)3 is 365 g/mol

Concentration for first trial = 0.1 x 1000 / 365 x 120 = 2.28 x 10-3 M

Hence, Ksp =(2.28 x 10-3)^3 ( 2.28 x 10-3) ^2 = 6.16 x 10-14 M^5

For second, total volume =100 + 5 mL = 105 mL

Molar mass of Ca2(PO4)3 is 365 g/mol

Concentration for 2nd trial = 0.1 x 1000 / 365 x 105 = 2.61 x 10-3 M

Hence, Ksp =(2.61 x 10-3)^3 ( 2.61 x 10-3) ^2 = 1.21 x 10-13 M^5

For 3rd trial, total volume =100 + 1 mL = 101 mL

Molar mass of Ca2(PO4)3 is 365 g/mol

Concentration for first trial = 0.1 x 1000 / 365 x 101 = 2.71 x 10-3 M

Hence, Ksp =(2.71 x 10-3)^3 ( 2.71 x 10-3) ^2 = 1.46 x 10-13 M^5

For 4th trial, total volume =100 + 7 mL = 107 mL

Molar mass of Ca2(PO4)3 is 365 g/mol

Concentration for first trial = 0.1 x 1000 / 365 x 107 = 2.56 x 10-3 M

Hence, Ksp =(2.56 x 10-3)^3 ( 2.56 x 10-3) ^2 = 1.10 x 10-13 M^5

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