For this problem ignore the basicity of phosphate ion (assume remains as PO 4 3-
ID: 513408 • Letter: F
Question
For this problem ignore the basicity of phosphate ion (assume remains as PO43- ).
a) The solubility of strontium phosphate in water is 0.14 mg per liter. What is the numerical value of Ksp for strontium phosphate? a) Ksp =__________
b) What is the solubility (in mg per L) of strontium phosphate in a 0.10 M solution of strontium acetate? b) sol. in M =________
c) Why is the answer to (b) different from the value given in (a)? _______________________________________________________________________
Explanation / Answer
Answer:
The solubility of strontium phosphate in water is 0.14 mg per liter.
Sr3(PO4)2= 0.14 mg = 0.00014 g
Molar mass Sr3(PO4)2= 452.8
1st :Write the balanced equation for the salt dissolving in water:
Sr3(PO4)2 (s) ó 3Sr+2(aq) + 2PO43-(aq)
2nd: Calculate the moles of salt:
Moles= Weight of chemical/ molar mass of chemical
=0.00014/452.8
Moles =3.09 X 10-7
3rd: Use reaction stoichiometry to determine mol Sr+2 and mol PO43-
3.09 X 10-7 Sr3(PO4)2 = 9.27 X 10-7 MolSr2+
3.09 X 10-7 Sr3(PO4)2 = 6.18 X 10-7 MolPO3-
Because there is exactly 1 L of solution, these values are also the molarities of each ion.
Ksp = [Sr2+]3[PO43-]2
= [9.27 X 10-7]3[ 6.18 X 10-7]2
= 3.04 x 10-31
b. What is the solubility (in mg per L) of strontium phosphate in a 0.10 M solution of strontium acetate?
Answer:
Sr3(PO4)2 (s) ó 3Sr+2(aq) + 2PO43-(aq)
3X+ 0.10 2X
Ksp = [Sr2+]3[PO43-]2
3.04 x 10-31 = [3X]3[2X]2
3.04 x 10-31 =(27 X3 )(8X)2
3.04 x 10-31=216X5
X= 2.68 X 10-7
Common ion Sr2+ = 3X+ 0.10
= 3 X 2.68 X 10-7 + 0.10
b= 0.1 M
c. Why is the answer to (b) different from the value given in (a)?
Because in case of b the extra common ion is added (i.e.Sr2+)
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