Determine the hydronium ion concentration and the pH of solution that is 0.10 M

Question

Determine the hydronium ion concentration and the pH of solution that is 0.10 M NaH_2PO_4 and 0.15 M Na_2HPO_4 (the K_a for H_2PO^-_4 is 6.2 times 10^-8). Include a chemical equation showing the relevant equilibrium. Finish the chemical equation showing the solubility equilibrium for calcium phosphate. Calculate the molar solubility of Ca_3(PO_4)_2 in a 0.20 M Na_3PO_4 solution. As part of your answer, fill in the grid with the relevant numbers or expressions. (K_sp for calcium phosphate is 1.3 times10^-32)

Explanation / Answer

Ksp = 1.3 x 10-32

Ca3(PO4)2 <-----> 3Ca2+ + 2PO43-

Initial: C 0 0 C= amount of the salt taken

Change -S + 3S +2S where S is the molar solubility of the salt

Equilibrium: C - S +3S +2S

Na3(PO4) <-----> 3Na+ + PO43-

Initial: 0.20 M 0 0

E: 0 3 x 0.2 M 0.2 M

[ PO43-] = 2S+ 0.2 ; [Ca2+] = 3S

Ksp Ca3(PO4)2 = [Ca2+]3 [ PO43-]2

1.3 x 10-32 = (3S)3 ( 2S + 0.20)2

   = 27S3 x (0.20)2 2S+ 0.20 = 0.20 since S<<<<<0.20M

= 1.08 S3

S = (1.3 x 10-32/1.08)1/3 = ( 1.20 x 10-32)1/3 = 2.3 x 10-11 M

The molar solubility of Ca3(PO4)2 in 0.2 M Na3 PO4 is 2.3 x 10-11 M

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