What minimum number of moles of NH4Cl must be added to 1.0 liter of solution that is 0.10 M in Mg(NO3)2 and 1.0 M in NH3 to prevent precipitation of Mg(No3)2?
Mg(OH)2 ==> Mg^+2 + 2OH^-
Ksp = (Mg^+2)(OH^-)^2
Kb for NH3 = (NH4^+)(OH^-)/(NH3)
Solve Ksp for OH^-. You now Mg so there is only one unknown. Substitute OH from that calculation into Kb for NH3. You know NH3 and Mg^+2, solve for NH4^+, then convert that molarity to grams NH4Cl. Post your work if you get stuck.
how do you know Mg?
To determine the minimum number of moles of NH4Cl needed to prevent precipitation of Mg(NO3)2 from the solution, we can use the concept of the common ion effect. The common ion effect states that the presence of a common ion in a solution reduces the solubility of a compound.
In this case, NH4Cl will provide the common ion NH4+ which will react with the NO3- ions from Mg(NO3)2, shifting the equilibrium towards the formation of MgNH4NO3(s) precipitate.
First, let's write the balanced equation for the reaction between NH4Cl and Mg(NO3)2:
Mg(NO3)2 + 2NH4Cl --> MgNH4NO3(s) + 2HCl
From the balanced equation, we can see that 1 mole of Mg(NO3)2 reacts with 2 moles of NH4Cl to form 1 mole of MgNH4NO3(s).
Given that the solution is 0.10 M in Mg(NO3)2, it means that there is 0.10 moles of Mg(NO3)2 in 1.0 liter of the solution.
Since 1 mole of Mg(NO3)2 reacts with 2 moles of NH4Cl, we multiply the moles of Mg(NO3)2 by 2 to find the minimum moles of NH4Cl needed:
0.10 moles Mg(NO3)2 × 2 moles NH4Cl/1 mole Mg(NO3)2 = 0.20 moles NH4Cl
Therefore, the minimum number of moles of NH4Cl that must be added to the solution is 0.20 moles.