Question

28. -70.1 points 0/4 Submissions Used (a) If the molar solubility of YF3 at 25 °C is 4.23e-06 mol/L, what is the Ksp at this temperature? Ksp =D (b) It is found that 0.0628 g of BaF2 dissolves per 100 mL of aqueous solution at 25 °C. Calculate the solubility-product constant for BaF2. Ksp = (c) The Ksp of Ca3(PO4)2 at 25 °C is 2.05e-35. What is the molar solubility of Ca3(PO4)2? solubility = mol/L

Answer 1

Solution (a) Ya (S) Y Coany 3f (a) Soeabieity: S S 35 Since, Ksp (3s) a3 = 0 (2s3) 2S KsP -6 Civen S= 4.23 X10meeL 4 ksP= 2)4.23 X10 2(s.20 o, -22 XID AMS -21 8.64 X1O YSP

(b) Civen = 0.06203 BaF2 of mass 100 Lite Velume af saeution100me 1000 20.100 L molar mass of Bat (2Xmo lor mas of BaF2 Molealor m46 (2X 19 13 + 1759/moe Given mars Of Baf males 0t BaF2= of number moleculor mas f Bof2

Oi0628 Baf2 1a5 /mae beaf2 3.6 XID male Soubiait DO Of mals Of Baf2 Ltres saeution 4 3.6 XIDM0 le 0.100 2e - 3.6X103 mae L Bat2(ag) + 2FCe) Baf26) Solabiety 2 S Since, C8a2 F Ksp = KsP 453 aut10)3 4X KSP 1 4 X(4.666 IO 7 1.87 XID KsP ARS

C -3 Cas (Poa)2 3cat2+ 2 Pa (04) Solau biity: S 2S 3 s Since, ksp Cate)3 = Cas 3S 23 l08s5 Sp -35 2.05 X1O Civen- KsP 5 108S 36 2.05 XID So, -35 2.05 X10 108 35 2.05 Y10 108 4.53XIDmad/As S moefL 4 53 X10