Hello I need help with this excersise: How many moles of NH4Cl are neccesary to add to 1 liter of solution of Co2+ 0,20 M in order to prevent the precipitation when the solution is saturated with H2S (0,1M) and the pH is 7,50. It 's known that pKbNH3=4,75 ; Kps(CoS)=2,0×10^-25 ; Ka1 (H2S)= 1,1×10^-7 ; ka2(H2S)= 10^-14 and Kc of Co (NH3)6=10^35,1
The reaction between NH4Cl and Co2+ is :
Co +6NH3 <----------->Co (NH3)6
I have calculated:
[S-2]= 1,1 ×10^-7 M
[Co+2] = 1,82 ×10^-18M
[NH3]=4,54×10^-4 M
[NH4]= 0,2551 M
Well I don't understand what to do next, in other web pages told me to do an addition:
[NH4Cl]= [NH3]+6 [Co (NH3)6]+[NH4] and then calculate the moles.
But I don't get why I have to do this addition to obtain the molarity of the solution of NH4Cl and why they put a 6 before the Co (NH3)6 because in the equation the coefficient of Co (NH3) is 1
That [NH4Cl] = [NH3] + [NH4^+] + 6[Co(NH3)6]^2+ is a mass balance equation. You are adding all of the species in which the NH4Cl will be. You multiply that complex by 6 because it contains 6 times the NH3 found in the other species. So you add all of them together to get the total NH4Cl.
To answer this question, we need to consider the equilibrium between Co2+ and NH3 in the given reaction: Co + 6NH3 ⇌ Co(NH3)6.
First, let's analyze the given information and calculate the concentrations of the relevant species.
Given:
[H2S] = 0.1 M (saturated solution)
pH = 7.50
pKbNH3 = 4.75
Kps (CoS) = 2.0 × 10^-25
Ka1 (H2S) = 1.1 × 10^-7
Ka2 (H2S) = 10^-14
Kc (Co(NH3)6) = 10^35.1
From the given data:
[S-2] = 1.1 × 10^-7 M (concentration of S^2- ions from H2S)
[Co+2] = 1.82 × 10^-18 M (concentration of Co2+ ions)
[NH3] = 4.54 × 10^-4 M (concentration of NH3)
[NH4] = 0.2551 M (concentration of NH4+ ions)
Now, let's consider the formation constant of Co(NH3)6 (Kf = Kc/[NH3]^6):
Kf = 10^35.1 / (4.54 × 10^-4)^6
Next, we need to find the molarity of NH4Cl needed to prevent precipitation. The idea here is to calculate the concentration of NH4+ ions and compare it to the concentration of S^2- ions from H2S. If the concentration of NH4+ is greater, it will prevent the precipitation of CoS.
To calculate the molarity of NH4Cl, let's consider the dissociation reaction of NH4Cl in water:
NH4Cl ⇌ NH4+ + Cl-
We can assume that all of the NH4Cl will dissociate, meaning the concentration of NH4+ ions is equal to the concentration of NH4Cl.
[NH4Cl] = [NH4]
Now, since you already have [NH4] = 0.2551 M, that is the concentration of NH4Cl needed to prevent precipitation.
To calculate the number of moles of NH4Cl needed, you can simply multiply the molarity by the volume (in liters) of the solution. In this case, you mentioned 1 liter of solution, so the number of moles of NH4Cl is 0.2551 mol.
So, to summarize:
The molarity of NH4Cl required to prevent precipitation is 0.2551 M, and the number of moles of NH4Cl needed is 0.2551 mol.
I hope this explanation helps you understand the calculation process and reasoning behind it.