Ionic Equation

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Tylor McGrew 1J
Posts: 20
Joined: Fri Sep 29, 2017 7:03 am

Ionic Equation

Postby Tylor McGrew 1J » Mon Dec 04, 2017 12:41 am

When writing the ionic equation, how does one know which molecules get split and which do not. For instance, why is (COOH) or H2O not split into different parts but NaOH is?

Diane Bui 2J
Posts: 61
Joined: Sat Jul 22, 2017 3:00 am

Re: Ionic Equation

Postby Diane Bui 2J » Mon Dec 04, 2017 1:10 am

In the example you gave, NaOH is an ionic compound and dissociates when it is in water because of hydrogen's and oxygen's partial charges that attract the ions in NaOH, causing it to dissociate into two separate ions. The hydrogen atom has a slightly positive charge, and the oxygen atom has a slightly negative charge. On the other hand, ionic compounds are crystals that have equal numbers of positive and negative ions. When put in water, the water molecules start to attract the ions. This is also why strong acids (i.e. HCL, HBr) and strong bases (i.e. NaOH, CaO) completely dissociate in water.

soniatripathy
Posts: 50
Joined: Thu Jul 13, 2017 3:00 am

Re: Ionic Equation

Postby soniatripathy » Mon Dec 04, 2017 11:38 am

The question is not super clear but if you are referring to writing net ionic equations then you know which species dissociates based on solubility rules and strong acids and bases. There are 3 basic steps to writing net ionic equations: (1) write a balanced equation with the states of matter included as well (2) use solubility rules and rules about strong acids and bases to write an ionic equations to show which species dissociates (3) cancel out the spectator ions to get the net ionic equation. Here is an example:

2 Na3PO4 (aq) + 3 CaCl2 (aq) --> 6 NaCl (aq) + Ca3(PO4)2 (s)
based on solubility we know that Na3PO4 and CaCl2 and NaCl dissociate into ions
6 Na+ (aq) + 2 PO43- (aq) + 3 Ca2+ (aq) + 6 Cl- (aq) --> 6 Na+ (aq) + 6 Cl- (aq) + Ca3(PO4)2 (s)
cancel out the spectator ions: 6 Na+ and 6 Cl-
you are then left with the net ionic equation of
2 PO43- (aq) + 3 Ca2+ (aq) --> Ca3(PO4)2 (s)

Clarisse Wikstrom 1H
Posts: 63
Joined: Fri Sep 29, 2017 7:05 am

Re: Ionic Equation

Postby Clarisse Wikstrom 1H » Mon Dec 04, 2017 5:18 pm

soniatripathy wrote:The question is not super clear but if you are referring to writing net ionic equations then you know which species dissociates based on solubility rules and strong acids and bases. There are 3 basic steps to writing net ionic equations: (1) write a balanced equation with the states of matter included as well (2) use solubility rules and rules about strong acids and bases to write an ionic equations to show which species dissociates (3) cancel out the spectator ions to get the net ionic equation. Here is an example:

2 Na3PO4 (aq) + 3 CaCl2 (aq) --> 6 NaCl (aq) + Ca3(PO4)2 (s)
based on solubility we know that Na3PO4 and CaCl2 and NaCl dissociate into ions
6 Na+ (aq) + 2 PO43- (aq) + 3 Ca2+ (aq) + 6 Cl- (aq) --> 6 Na+ (aq) + 6 Cl- (aq) + Ca3(PO4)2 (s)
cancel out the spectator ions: 6 Na+ and 6 Cl-
you are then left with the net ionic equation of
2 PO43- (aq) + 3 Ca2+ (aq) --> Ca3(PO4)2 (s)


Just to clarify, HI(aq) WOULD dissociate in water b/c it is a strong acid. However, would HI(g)? No, correct, since it is a gas?


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