I have this exercise in my chemistry book, and I have been staring at it for so long but unfortunately I have no single idea what to do and where to start. I will be so thankful for any help.
$0.05$ moles of NH3 are added to one liter of MgCl2 of concentration $0.02$ mol/L. How many moles of NH4Cl do we need to add in order to prevent the precipitation of Mg(OH)2? Given Ksp=8.4×10^(-12) ; pKb=4.5
My teacher's approach:
The solution of Mg(OH)2 is saturated at pH=10.4 since: Ksp=4s³ Then s=1.28×10^(-4) mol/L with [HO–] = 2s
Afterwards, using Henderson–Hasselbalch equation, using the given and calculated values, we get
I thought that the calculated value of $s$ is true when Mg(OH)2 is put in pure water and not in a solution containing a weak base which decreases its solubility by common ion effect, so this method is not valid. But i don't know how to solve the problem. Can you help me please? Sorry for the not perfect format. I did my best.