Solid Ca3(PO4)2 is placed into 10.0 L of water.When equilibrium is established, the concentration of Ca2+ is 2.3x10-4 M. What is Kc for this equilibrium?

Answer 1

3.7*10^-15

The balanced reaction for this problem is:

Ca₃(PO₄)₂ (s) ⇔ 3Ca²⁺ (aq) + 2PO₄^3- (aq)

If [Ca²⁺]=2.3*10^-4M, then we already have all the values we need for this reaction.

The molar ratio here for [Ca₃(PO₄)₂] : [Ca²⁺] : [PO₄^3-] is 1:3:2 based on their coefficients.

So, [Ca₃(PO₄)₂]= [Ca²⁺]/3 = 7.667*10^-5 M,

[Ca²⁺]=2.3*10^-4M as given, and

[PO₄^3-]= [Ca²⁺]*2/3 = 1.5333*10^-4M.

We use the formula Kc= ([B]^b[C]^c)/([A]^a) for the reaction aA ⇔ bB + cC.

In this reaction, it would be Kc=([Ca²⁺]³[PO₄^3-]²)/([Ca₃(PO₄)₂]¹).

Plugging in the numbers above, we get Kc=((2.3*10^-4)³(1.5333*10^-4)²)/(7.667*10^-5), which equals 3.731*10^-15. Since we are given 2 significant figures in the question, our final answer is 3.7*10^-15.

Please let me know if this helps or if you are still confused about a part of this!