Calculate the molar solubility of Zinc II Phosphate(Zn3(PO4)2). Ksp = 9.0e-33. ?

1 Answer
Truong-Son N.
Apr 1, 2018

#s = 1.53 xx 10^(-7) "M"#

How will the molar solubility change as #"pH"# decreases? Why?

The molar solubility is the same as the concentration of the ion with a coefficient of #1#... thus, we designate that #s#, and write the reaction as:

#"Zn"_3("PO"_4)_2(s) rightleftharpoons 3"Zn"^(2+)(aq) + 2"PO"_4^(3-)(aq)#

#"I"" "" "-" "" "" "" "" "0" "" "" "" "" "0#
#"C"" "" "-" "" "" "" "+3s" "" "" "+2s#
#"E"" "" "-" "" "" "" "" "3s" "" "" "" "2s#

And lest we forget, the coefficient also goes into the exponent.

#K_(sp) = 9.0 xx 10^(-33) = ["Zn"^(2+)]^3["PO"_4^(3-)]^2#

#= (3s)^3(2s)^2#

#= 27s^3 cdot 4s^2#

#= 108s^5#

And so, the molar solubility is:

#color(blue)(s = (K_(sp)/108)^(1//5) = 1.53 xx 10^(-7) "M")#

What is the molar concentration of #"PO"_4^(3-)#?

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