# Using a weak acid to prevent the precipitation of Mg(OH)2

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

pH=14+log[HO]

Afterwards, using Henderson–Hasselbalch equation, using the given and calculated values, we get

[NH4+]=6.29×10^-3 mol/L

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.

• It is unclear to what pKb refers to. As pKb1+pKb2 of Mg(OH)2 is 1.8 resp 2.56 ( periodensystem-online.de/… ), it must be pKb of NH3, which is 4.75, so 4.5 is quite incorrect. – Poutnik Sep 13 at 23:18

Consider (1) for $$\ce{Mg(OH)2}$$ dissolution equilibrium, (2) for ammonia acido-basic equilibrium.
Combining both, you get required concentration of $$\ce{NH4+}$$.
\begin{align} \mathrm{p}K_\mathrm{sp} &= \mathrm{pMg} + 2 \cdot \mathrm{pOH}\tag{1}\\ \mathrm{pOH} &= \mathrm{p}K_\mathrm{b, NH3} + \mathrm{pNH3} - \mathrm{pNH4+} \tag{2}\\ \mathrm{p}K_\mathrm{sp} &= \mathrm{pMg} + 2 \cdot \left( \mathrm{p}K_\mathrm{b, NH3}+ \mathrm{pNH3} - \mathrm{pNH4+}\right) \tag{3}\\ \mathrm{pNH4+} &= \frac 12 \left( \mathrm{pMg} - \mathrm{p}K_\mathrm{sp} \right) + \mathrm{p}K_\mathrm{b, NH3}+ \mathrm{pNH3} \tag{4}\\ \end{align}
where "$$\mathrm{pX}$$" means "$$-\log{X}$$" resp. "$$-\log{c_\mathrm{X}}$$".