Question

Write a balanced net ionic equation to show why the solubility of Ca3(PO4)2 (s) increases in the presence of a strong acid and calculate the equilibrium constant for the reaction of this sparingly soluble salt with acid. Consider only the FIRST STEP in the reaction with strong acid. Use the pull-down boxes to specify states such as (aq) or (s). Caş(PO4)2 / 3 + 2H20* 60 3 3ca?+ 9 3 + 2HPO3* _q 3 + 24, 0 ° K= 0.22x10^-12 Submit Answer Retry Entire Group No more group attempts remain

the answer i have entered is incorrect

Ksp = 1.0 x 10^-25

Ka =

Phosphoric acid

H3PO4

7.5×10-3

6.2×10-8

3.6×10-13

Answer 1

Solution-

The balanced net ionic equation can be written as below :

Ca3(PO4)2 (s) + 2 H3O+ (aq) 3 Ca2+ (aq) + 2 HPO42- (aq) + 2 H2O (l)

Now let's find the value of K
We have

Reaction 1 : Ca3(PO4)2 (s) 3 Ca2+ (aq) + 2 PO43- (aq) : here K1 = Ksp = 1.0 x 10-25

Reaction 2 : 2 PO43- (aq) + 2 H2O (l) 2 HPO42- (aq) + 2 OH- (aq) => K2 = 7.716 x 10-4

Reaction 3 : 2 H3O+ (aq) + 2 OH- (aq) 4 H2O (l) => K3 = 1 x 1028

adding all the three reactions we get the balanced net ionic equation

K = (K1) * (K2) * (K3)

K = (1.0 * 10^-25) * (7.716 * 10^-4) * (1* 10^28)

Hence we get

K = 0.772