# Can i calculate the solution of the problem in more simple way?

and i got a serious question. During my exercises at home, i found that im struggling a lot while solving simple problem`s like this one:

I have an aquatic substance with 0.100 M of $\ce{Mg(NO3)2}$ and 0.100 M of $\ce{NH3}$. Given that: $K_\mathrm{sp}(\ce{Mg(OH)2})=1.50\times10^{-11}$ and $K_\mathrm b(\ce{NH3})=1.75\times10^{-5}$

I need to find:

1. If there will be any precipitation for $\ce{Mg(OH)2}$?

2. What concentration of ammonia I need to begin the precipitation of $\ce{Mg(OH)2}$ in a solution of 0.100 M $\ce{Mg(NO3)2}$?

3. If $\ce{Mg(OH)2}$ was a soluble salt, what was the pH in the solution?

And the process of solving this takes me a lot of time. What I did to solve are those steps:

a. To decide if there will me a percipitation I need to know my Q of magnesium hydroxide and compare it to K. Which will give me a "True" for a precipitation to occur.

b. Then for the next question i made an ICE table where X was the initial conc. of ammonia and the other fields were dependent on the hydroxide concentration from the previous percipiation(stochiometry). This step was long and its included about 2–3 tables.

c. If magnesium hydroxide was a soluble salt, probably he would not have an equilibrium constant and will dissolve in one direction reaction. That tells me about the dependence of the pH in solution, which is based on $\ce {OH^-}$ concentration.

It would be lovely for some of you who know much better than i do, to correct me or show me much simpler ways to think outside the box. Maybe it will reduce my amount of time spent on those exercise (in the best case scenario), if not - it sure will help me or somebody else in other ways. Thanks for support.

• I don't think that there is any sword to cut cleanly through this Gordian knot, so don't waste a lot of time looking for one. Note however that solving either 2 or 3 gives you 1.
– MaxW
Dec 30 '17 at 20:07

I think you are overthinking this.

Using the solubility product and the nominal $\ce{Mg^2+}$ concentration, calculate the $[\ce{OH-}]$ concentration that must be exceeded to cause precipitation of the magnesium hydroxide:

$$[\ce{OH-}]>\sqrt{\frac{K_{\mathrm{sp}}}{C(\ce{Mg^2+})}}$$

Now use the base constant equilibrium constant (in its simplified form), to calculate the actual $[\ce{OH-}]$ concentration, due to the ammonia present:

$$K_\mathrm b(\ce{NH3})\approx \frac{[\ce{OH-}]^2}{C(\ce{NH3})}$$

$$[\ce{OH-}]\approx \sqrt{K_\mathrm b(\ce{NH3})\times C(\ce{NH3})}$$

If the latter exceeds the former, precipitation will occur.

And as @MaxW noted (in the comments):

Note however that solving either 2 or 3 gives you 1.

For $2$, use the calculated $[\ce{OH-}]$ concentration required for precipitation to occur, to calculate the minimum $C(\ce{NH3})$, using:

$$C(\ce{NH3})\approx \frac{[\ce{OH-}]^2}{K_\mathrm b(\ce{NH3})}$$

• Thanks a lot. Your answer definitely organized my thoughts. Dec 31 '17 at 20:22