Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

omogeneous Precipitation recipitation in Presence of Electrolyte dsorption, Abso

ID: 717834 • Letter: O

Question

omogeneous Precipitation recipitation in Presence of Electrolyte dsorption, Absorption, Inclusion, Occlusion eprecipitation, Coprecipitation OMEWORKt13DUE9/10/18: 1. From the equations HFH+F NHNH,H K, 6.8.10 K, 5.69 10t K,-? Find the value for Ky Does this equilibrium lie on the left or on the right site? 2. A water sample of 1 L volume is contaminated with 862 ppm Pb. How many grams of Nal do you need to add to precipitate 99.99% of the lead in this water sample by the following precipitation reaction? " + 21-_ Pbla To answer the question: a) Calculate the ppm, mg, and mol/L of Pb left in solution. b) Write the K expression for Pbl, and find the iodide concentration [Th Q left in solution) using the [Pb' calculated in a). c) Calculate the mass of Nal using the iodide concentration calculated in b). d) Calculate the mass of Nal you need to make up the Pblao (99.99%). 3. A solution contains 0.005 M of each of the following anions: Ca", La", Ag'. If this solution is treated with CO, 2, in what order will the cations precipitate? 4. Explain the terms adsorption, absorption, inclusion, and occlusion.

Explanation / Answer

Ans 1

The first reaction

HF = H+ + F-

K1 = 6.8 x 10^-4

The second reaction

NH4+ = NH3 + H+

K2 = 5.69 x 10^-10

Reverse the reaction

K2'' = 1/K2 = 1/(5.69 x 10^-10) = 1.76 x 10^9

Now add the two reactions

NH3 + HF = NH4+ + F-

K3 = K1 x K2''

= 6.8 x 10^-4 x 1.76 x 10^9

= 1.196 x 10^6

K >> 1

The equilibrium lie on the right side