Question

Solubility Product Constants (Ksp at 25 °C) Type Formula Cus 7.9 x 10-37 Fes 4.9 x 10-18 Fe2S3 1.4 x 10-88 Solubility Product Constants (Ksp at 25 °C) Type Formula BSP Zn(OH)2 4.5 * 10-17 Iodides Pb12 8.7 x 10-9 Agi 1.5 x 10-16 Type Formula MnCO3 1.8 * 10-11 NiCO3 6.6 x 10 Ag2CO3 8.1 x 10-12 ZnCO3 1.5 * 10-11 Chlorides PbCl2 1.7 x 10-5

1)

Use the References to access important values if needed for this question.

The maximum amount of lead chloride that will dissolve in a 0.138 M ammonium chloride solution is M.

2)

Use the References to access important values if needed for this question.

The molar solubility of lead iodide in a 0.138 M ammonium iodide solution is M.

3)

Use the References to access important values if needed for this question.

The molar solubility of iron(III) sulfide in a 0.267 M iron(III) nitrate solution is M.

Answer 1

Answer:-

These questions are answered by using simple concept of calculation of solubility using the expression of Ksp and its value.

The answer is given in the image,

Answer:- 1) Pbalacs) = Abiton + 2Cléag, Mspa 1.7x10-5 0.138M Asp= [Pb2+] [ce]2 1-7x10-5=(P621] (0-L38)2 47x10-5=1Pb287 (0-138)2 (Pb2+] = 8.93x10-4M LP bdą] = 8.93x10-4M » Pb1zon = Pb mees +21cm) KIT sp = 8.7x1009 0.138 [MH41] = 0·138M=[1] Asp = [Pb2+] [172 8.7x10-9 - [Pb2+] (0-13812 [Pb2+] =4.57X107M [Pb Ig] =4.57X107 M 3) Fez Szcs) 2F22key+ 3 slags 0.267 354) Ksp = 1.4x10-88 Esp = [Fe2+]²[5273 1.4X1088 = (0.267)?(85)3 1.4x10-68–60.26733x2753 53-7.27x10-89 S = 4.17 X10-30M Molar folubility = 4.17 X10-30M