Dissociation of calcium phosphate take place as follows-
Ca3(PO4)2 --------> 3Ca2+ + 2PO43-
Ksp = solubility product constant.
Solubility product expression for above reaction can be written as-
Ksp = [Ca2+]3 [PO43-]2
This is the correct expression for solubility product.
Hence correct option is (b).
Question 24 1 pts Write the solubility product constant expression for calcium phosphate, Ca3(PO4)2 O [Ca2+][PO43-M[Ca3(PO4)2]...
If the concentration of phosphate ions ([PO43-1) in a saturated solution of Ca3(PO4)2 is known, which expression should be used to calculate the solubility product constant (Ksp) for Ca3(PO4)2? Kp - 8 [PO43-15/27 Ksp - 3(PO4"-1/2 Ksp = 27 (PO43-1978 Ksp -9 [PO43-1514 Kup 4(PO43-119
Select the correct solubility equilibrium for the slightly soluble salt, calcium phosphate. O CaPO4(s) -- Ca2+(aq) + PO42-(aq) O Ca3(PO4)2(s) — 3Ca2+(aq) + 2P043-(aq) O Ca3PO4(s) — 3Ca+(aq) + PO43-(aq) CaPO4(s) - Ca(aq) + PO4(aq)
what is the molar solubility of Ca2+ in a 1.00 M aqueous solution of Ca3(PO4)2 (Ksp for calcium phosphate is 2.0 x 10^-29)
Given that the solubility reaction for calcium phosphate is Ca3(PO4)2(s) = 3Ca2+ (aq) + 2PO43- (aq) why does the addition of acid increase the solubility of calcium phosphate? View Available Hint(s) O It decreases the phosphate ion concentration, forcing the equilibrium to the right It decreases the phosphate ion concentration, forcing the equilibrium to the left. O It increases the phosphate ion concentration, forcing the equilibrium to the right. It increases the phosphate ion concentration, forcing equilibrium to the left....
A chemist fills a reaction vessel with 0.980 g calcium phosphate (Ca3(PO4)2) solid, 0.212 M calcium (Ca+2) aqueous solution and 0.119 M phosphate (PO4−3) aqueous solution at a temperature of 25.0°C. Under these conditions, calculate the reaction free energy ΔG for the following chemical reaction: Ca3(PO4)2(s) 3Ca+2(aq)+2PO4−3(aq) Use the thermodynamic information in the ALEKS Data tab. Round your answer to the nearest kilojoule.
Show calculations for Calcium phosphate molar solubility. Ca3(PO4)2. ksp = 2.08 x 10^-33. Can you please show me how you get from one step to the next? thanks!
Calcium phosphate (Ca3(PO4)2) is sparingly soluble in water. If the concentration of calcium ions in solution at equilibrium is 1.0 x 10°M, what is the Ksp of calcium phosphate? a. 4.4 x 10-46 b. 2.2 x 10-46 c. 1.1 x 10-46 d.5.5 x 10-47 a. b. O C. d. KESCION 20 5 pts 14a) Sodium chromate is added to a solution of 0.0060 M 5r2+ What is Q if the final concentration of Cro 2 is 0.0030 M? a. 7.2...
Ogen O Combi 7.) Consider the following reversible process in a closed system: ht ausw Ca3(PO4)2 (s) $ 3 Cat2 (aq) + 2 P04 (aq) Ke=2.0 x 10-29 Calculate the equilibrium concentrations of both calcium and phosphate ions in a saturated solution of calcium phosphate. (Ans. 2.14 x 106 M Ca2 and 1.4 x 106 M Pi)
10. Magnesium phosphate, Mg3(PO4)2, has a solubility product constant of 2.27 x 103. a. What is the molar solubility of Mg3(PO4)2 in 0.925 M potassium phosphate, K3PO4? (6 points) b. What is the molar solubility of Mg3(PO4)2 in pure water? (5 points)
Phosphoric acid can be prepared from calcium phosphate according to the following reaction: Ca3(PO4)2 (s) + 3 H2SO4 (l) → 3 CaSO4 (s) + 2 H3PO4 (l) 310. g/mol 98.0 g/mol 136 g/mol 98.0 g/mol [a] If 0.664 mole of Ca3(PO4)2 are combined with 1.53 mole of H2SO4, how many grams of H3PO4 can be produced? [4 pts] [b] If 62.6 g of H3PO4 is actually produced in the above reaction, what is the percent yield for the reaction? [2...