Geniuses An aqueous solution it prepared by dissolving 61.5 milligrams of Ca(NO_

Question

Geniuses An aqueous solution it prepared by dissolving 61.5 milligrams of Ca(NO_3)_2 (a highly soluble salt) in 5 00L of solution. Assuming the Ca(NO_3)_3 is completely dissolved, find the molar concentration of Ca^2+ ions (M) in the solution Na_3PO_s [i.e phosphate ion] is slowly added to the solution in part a, until precipitation of Ca_3 [PO_4)_4 begins. Given that K_sp for Ca_3 (PO_4)_2 = 1.0 Times 10^-14, find the molar concentration of PO_4^3- (M) at which precipitation of Ca_3(PO_4)_2 will begin (If you couldn't solve part a, assume a FAKE initial [Ca^2+] = 1.25 Times 10^-4 M Using your answer to part b, find the maximum number of milligrams of Na_3PO_4 that can be added to the solution before precipitation of Ca_3 (PO_4)_2 begins. Recall that the volume of the solution is 5.00 L (If you couldn't solve part b, assume a FAKE [PO_4^3-] = 2.62 Times 10^-6 M)

Explanation / Answer

a)
Molar mass of Ca(NO3)2 = 164.1 g/mol
Number of moles of Ca(NO3)2 = mass/molar mass
= (0.0615)/164.1
= 3.75*10^-4 mol
V = 5 L

[Ca(NO3)2] = number of moles/volume
= (3.75*10^-4)/5
=7.5*10^-5 M

[Ca2+] = 7.5*10^-5 M

b)
Ca3(PO4)2 ---> 3Ca2+ + 2 PO43-
7.5*10^-5 2s

Ksp = [Ca2+]^3 * [PO43-]^2
1*10^-26 =(7.5*10^-5)^2 * (2s)^2
4*s^2 = 1.78*10^-18
s = 6.67*10^-10 M

[PO43-]=2s = 2*6.67*10^-10 = 1.334*10^-9 M
Answer: 1.334*10^-9 M M

c)
[PO43-] = 1.334*10^-9 M
3 Na+ is required for each PO43-

[Na3PO4] = 1.334*10^-9 M

Molar mass of Na3PO4 = 164 g/mol

[Na3PO4] = mass / (molar mass * volume)
1.334*10^-9 M = mass / (164*5)
mass = 1.09*10^-6 g
= 1.09*10^-3 mg
Answer: 1.09*10^-3 mg

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