Question

About what minimum mass (in mg) of solid NaF would have to be added to 1.00 L of 0.00010 M CaCl2 solution in order to begin to precipitate solid CaF2? (Assume constant volume during dissolution. Ksp of CaF2 = 4.0 x 10−11 at 25°C). (1) 13 mg (2) 17 mg (3) 63 mg (4) 32 mg (5) 27 mg

Formic acid occurs naturally in most ants. Soap solutions containing calcium hydroxide are used to treat ant bites forming calcium formate. Given Ka of HCOOH (formic acid) is 1.8 X 10-4 at room temperature, what answer is closest to the pH of the salt solution of 0.5 M calcium formate at room temperature? (1) 2.02 (2) 5.28 (3) 8.72 (4) 11.98 (5) None of the them (1-4)

The concentration of the complex ion in each of following solutions is 1.00 M. In which of the solutions will the concentration of the uncomplexed metal ion be the greatest? Hg(CN)4 2– (Kf = 9.3 x 1038) Be(OH)4 2– (Kf = 4.0 x 1018) Zn(OH)4 2– (Kf = 3.0 x 1015) Cu(NH3)4 2+( Kf = 5.6 x 1011) CdI4 2– (1.0 x 106 ) (1) Hg2+ (2) Be2+ (3) Zn2+ (4) Cu2+ (5) Cd2+

Answer 1

First Answer

In order to get the precipitation of CaF2, the ionic product should be greater than the solublity product. Given the value of the solubility product (K_SP)of CaF2 is 4.0*10-11.

where K_SP = [Ca²⁺] [F^-]²

the product of [Ca²⁺] [F^-]² should be the > ionic product.

[Ca²⁺] = 0.0001

the concentration of NaF for (1): 0.013/41.98 = 3.09*10^-4

FOR (2): 0.017/41.98 = 4.08*10^-4

FOR (3) 0.063/41.98 = 1.5 * 10^-3

FOR (4) 0.032/41.98 = 7.6*10^-4

FOR (5) 0.027/41.98 = 6.4*10^-4

(1 ) [Ca²⁺][F^-]² = (1*10^-4) (3.0*10^-4)² = 9*10^-12

(2) [Ca²⁺][F^-]² = (1*10^-4) (4.0*10^-4)² = 16*10^-12

(3) [Ca²⁺][F^-]² = (1*10^-4) (1.5*10^-3)² = 2.25*10^-10

(4) = 57.7*10^-12

(5) = 40.9*10^-12

^{AS WE CAN SEE ABOVE ONLY IN ONE CASE (IN 3 CASE) WE GOT IONIC PRODUCT GREATER THAN SOLUBILITY PRODUCT.}

ANS IS 63mg