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Altius COVID 3 questions
Natasha_4902
#1 Posted : Monday, July 13, 2020 4:35:26 AM
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Hi,

In the biology section of the Altius 3 COVID exam, for question 33, why isn't the equilibrium constant 1/((ca2+)^3(PO4)^2). I thought that phosphate precipitates Ca2+, so wouldn't the product be CaPO4, and Ca2+ and PO4 be the reactants (thus they would be in the denominator of the equilibrium constant)?

Thank you
INSTR_Katerina_102
#2 Posted : Monday, July 13, 2020 4:34:30 PM
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Hello,

The short answer is that you need to balance charge as Ca2+ is a 2+ cation and needs to form a neutral salt with the (PO4)3- anion. I have included a longer answer here (which includes why Ca and PO4 are not on the denominator):

For this question, a precipitation reaction is governed by a Ksp, or the equilibrium constant for precipitation. From this question, I would infer from the options given that the question is asking for me to find what the Ksp of this reaction is (as all of these look like Ksp expressions).

To get this Ksp, we need to write the precipitation reaction with Ca^2+ and PO4^2-.

The unbalanced equation would be as follows:

Ca^2+ + PO4^3- --> CaPO4

We need our salt to be overall neutral, so we need to multiply the charges by coefficients x and y to get the stoichiometry of the reaction, getting the equation:

xCa^2+ + yPO4^3- --> CaxPO4y

0 = 2x - 3y, where 2 and 3 are the charges on Ca and PO4 respectively, and x and y are stoichiometric coefficients.

We can then realize that x = 3 and y = 2 would be the lowest numbers satisfying this system, giving the balanced equation:

3Ca^2+ + 2PO4^3- --> Ca(3)(PO4)(2)

The Ksp stands for the solubility product and it describes the max amount of salt dissolved in solution, so we'll look at the reverse reaction (to skip this step, just remember the ion concentration is on top for Ksp).

Ca(3)(PO4)(2) --> 3Ca^2+ + 2PO4^2-

Ksp = [Ca^2+]^3[PO4^3-]^2

Please let me know if this is unclear, Ksp can be tough!
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