How do you determine precipitation between two compounds?

Question

for example:

KOH (aq) + MgBr2 (aq)

(NH4)2SO4 + ZnCl2

all numbers are subs.

I think one of them, no reaction takes place. Can you explain how this works!??

Answer 1

KOH (aq) + MgBr2 (aq) -->  ???

Knowing that this is a double replacement reaction, we can predict the products by "flip flopping" the positive and negative "ions".

KOH (aq) + MgBr2 (aq) ---> KBr + Mg(OH)2

Then check each of the predicted products in the solubility rules* to see if either is insoluble. KBr is soluble in water, but Mg(OH)2 is insoluble.  Therefore, we have a reaction because there is at least one insoluble product.  As so we write a balanced chemical equation.

2KOH (aq) + MgBr2 (aq) ---> 2KBr(aq) + Mg(OH)2(s)

--------------------------------------...

(NH4)2SO4 + ZnCl2 ---> ???

Predict the two products and check their solubilities in water.  Both are soluble.

(NH4)2SO4 + ZnCl2 ---> NH4Cl(aq) + ZnSO4(aq)

Therefore, there is no insoluble product, and no reaction

(NH4)2SO4 + ZnCl2 ---> no reaction

--------------------------------------...

*  If you need to look at the solubility rules, you can find them here

http://www.csudh.edu/oliver/chemdata/sol...

========== Follow up ==========

I just happened to notice that JohnDx5454 has misidentified the chemical family for ZnSO4.  ZnSO4 is zinc sulfate, and is soluble in water.  ZnSO4 is not zinc sulfide.  Zinc sulfide is ZnS and it is insoluble in water.

Answer 2

All compounds of the ammonium ion (NH4+), and of Alkali metal (Group IA) cations, are soluble.

# All nitrates and acetates (ethanoates) are soluble.

# All chlorides, bromides and iodides are soluble EXCEPT those of silver, lead and mercury(I).

# All sulphates are soluble EXCEPT those of silver, lead, mercury(I), barium, strontium and calcium.

# All carbonates, sulfites and phosphates are insoluble EXCEPT those of ammonium and Alkali metal (Group IA) cations.

# All hydroxides are insoluble EXCEPT those of ammonium, barium and alkali metal (Group I) cations.

# All sulfides are insoluble EXCEPT those of ammonium, Alkali metal (GroupI) cations and Alkali earth metal (Group II) cations.

# All oxides are insoluble EXCEPT those of calcium, barium and Alkali metal (Group I) cations; these soluble ones actually react with the water to form hydroxides (hydrolyse).

Answer 3

You need to use the solubility rules. The solubility rules will tell you which compounds are soluble and where compounds are insoluble. Then you can write the net ionic equations and complete net ionic equations, etc.

2KOH (aq) + MgBr2 (aq) --> 2KBr (aq)+ Mg(OH)2 (s)

according to the solubility rules all hydroxides are insoluble, but there are a few exceptions (group 1 elements, Sr and Ca)

2K+ (aq) + 2OH- (aq) + Mg2+(aq) + 2Br- (aq) --> 2K+(aq) + 2Br-(aq) + Mg(OH)2 (s)

Mg2+(aq) + 2OH- (aq) --> Mg(OH)2 (s)

(NH4)2SO4 (aq) + ZnCl2 (aq) --> ZnSO4(s) + 2NH4Cl (aq)

2NH4+ (aq) + SO4^-2 (aq) + Zn2+ (aq) + 2Cl- (aq) --> Zn2+ (aq) + SO4^-2 (aq) + 2NH4+ (aq) + 2Cl- (aq)

Edit:

Most "sulfates" are soluble. I was recalling the rules from memory and mixed up sulfides with sulfates. This is what the answer should be:

Zn2+ (aq) + SO4^-2 (aq) --> no reaction