Question

Based on the results from this experiment, predict whether some of the following combinations of compounds would precipitate or stay in solution. Explain how you determined that a certain compound should precipitate or be soluble. a. LiOH NaCl b. BaCl2 Na3PO4 c. MgSO4 KOH

Answer 1

Answer:

* No precipitate: $LiOH(aq)+NaCl(aq)\rightarrow LiCl(aq)+NaOH(aq)$

* Precipitate: $3BaCl_2(aq) +2Na_3PO_4(aq) \rightarrow Ba_3(PO_4)_2(s)+6NaCl(aq)$

* Precipitate: $MgSO_4(aq)+2KOH(aq)\rightarrow Mg(OH)_2(s)+K_2SO_4(aq)$

Explanation:

Hello!

In this case, since these all are double displacement reactions, in which the cations and anions are exchanged, we can write the resulting chemical reactions as follows:

a. LiOH and NaCl: No precipitate is formed since LiOH and NaOH are both largely soluble in water:

$LiOH(aq)+NaCl(aq)\rightarrow LiCl(aq)+NaOH(aq)$

b. BaCl2 and Na3PO4: barium phosphate precipitate is formed because it has a large molar mass which makes it insoluble in water:

$3BaCl_2(aq) +2Na_3PO_4(aq) \rightarrow Ba_3(PO_4)_2(s)+6NaCl(aq)$

c. MgSO4 and KOH: magnesium hydroxide "milky" precipitate is formed because it is not soluble in water:

$MgSO_4(aq)+2KOH(aq)\rightarrow Mg(OH)_2(s)+K_2SO_4(aq)$

Moreover, we can relate the solubility of a substance by considering its polarity, molar mass and nature; usually, heavy substances tend to be insoluble in water as well as nonpolar compounds.

Best regards!