I need help getting the net ionic equations for the equilibrium of sodium cyanide and water and calcium nitrate and water
Are these separate solution?
NaCN + H2O ==> Na^+(aq) + CN^-(aq)
To determine the net ionic equations for the equilibrium of sodium cyanide and water, and calcium nitrate and water, we first need to understand the dissociation behavior of each compound in water.
1) Sodium cyanide (NaCN) dissociates into sodium ions (Na+) and cyanide ions (CN-) in water:
NaCN (aq) --> Na+ (aq) + CN- (aq)
2) Water (H2O) itself dissociates slightly into hydronium ions (H3O+) and hydroxide ions (OH-):
H2O (l) --> H3O+ (aq) + OH- (aq)
Now, let's consider the equilibrium between sodium cyanide and water:
NaCN (aq) + H2O (l) ⇌ Na+ (aq) + CN- (aq) + H3O+ (aq)
The net ionic equation is obtained by removing the spectator ions, i.e., the ions that do not participate in the reaction. In this case, the sodium ion (Na+) is a spectator ion:
CN- (aq) + H2O (l) ⇌ CN- (aq) + H3O+ (aq)
Simplifying further, the net ionic equation for the equilibrium of sodium cyanide and water is:
H2O (l) ⇌ H3O+ (aq) + OH- (aq)
Now, let's look at the equilibrium between calcium nitrate and water:
Ca(NO3)2 (aq) + H2O (l) ⇌ Ca2+ (aq) + 2NO3- (aq) + 2H2O (l)
Again, the net ionic equation is obtained by removing the spectator ions, which in this case are the nitrate ions (NO3-).
Ca2+ (aq) + H2O (l) ⇌ Ca2+ (aq) + 2H2O (l)
Simplifying further, the net ionic equation for the equilibrium of calcium nitrate and water is:
H2O (l) ⇌ 2H2O (l)
Note that in this case, the net ionic equation appears to be a bit redundant, but it emphasizes the fact that water remains unchanged in the reaction.
I hope this helps! Let me know if you have any further questions.