Net ionic equation for Na₂SO₄ + Ba(NO₃)₂ precipitation

[PhysicsinCalifornia]

I just needed help understanding the concept of this chemical reaction

$$Na_2SO_4 (aq) + Ba(NO_3)_2 (aq) \rightarrow BaSO_4 (s) + 2NaNO_3 (aq)$$
This is the conventional equation, and I wrote the net ionic form to be:

$$SO_4^{2-} (aq) + Ba^{2+} (aq) \rightarrow BaSO_4 (s)$$

However, the answer in the back says:
$$[Ba \cdot NO_3]^+ (aq) + SO_4^{2-} (aq) \rightleftharpoons BaSO_4 (s) + NO_3^- (aq)$$

Can anyone tell my why the books answers it like that?

I did the experiment on this and the barium sulfate precipitate took several minutes to form (a slow reaction) Is that why?

 

[mrjeffy321]

According to my CRC Handbook of Chemistry and Physics,
the solubility product of Barium Nitrate (Ba(NO3)2) is about 4.64 E-3. This Ksp value is not exceptionally high, meaning that Ba(NO3) is a weak electrolyte.
This Ksp value corresponds to the degree of which Ba(NO3)2 will break into BaNO3+ and NO3- ions in solution,
Ba(NO3)2 (s) <===> BaNO3+ (aq) + NO3- (aq)

If the Ksp value for the neutral Ba(NO3) is so low, it would be expect to drop even lower (disassociate even less) when one is talking about the 2nd Nitrate ion disassociating,
BaNO3+ <===> Ba+2 (aq) + NO3- (aq)
Although I don’t have the value for this Ksp, it is probably much less than the first Ksp.


So to get back to your question,
your book may be assuming that the Ba(NO3)2 does not break up into Ba+2 ions to any significant quantity, but rather, stays as BaNO3+ after loosing the first Nitrate.