Net Ionic Equation: Ca^2+(aq) + 2H^+(aq) + SO_4^2-(aq) --> H_2O(l) + CaSO_4(s)

[Tom McCurdy]

I am supose to write the net ionic equation for

Complete the right side of the following molecular equation (but do not enter). Then enter the net ionic equation. Assume all salts formed are soluble. Acid salts are possible. Use = instead of ==>. Do not use spaces or subscripts. You may use brackets.


$$Ca(OH)_2(aq) + 2H_2SO_4(aq) ==>$$

I get that the products would be $$H_2O(l)+CaSO_4(s)$$

So Do I just have to balance this equation? then separate the two aq on the left into ions and leave the right as it is.

 

[GCT]

You should also consider the "acidic salt", this is sort of a weird question, because it neglects the effect of precipitation, if such a salt CaS04 precipitates, than it would drive the dissociation of HS04- forward, which usually doesn't occur to a great extent. But I think they just want you to play dumb on this one.

Once you find all of the salts, the net ionic equation will pertain to just these salts...for instance

$$Ca^{2+}_{(aq)} + SO_4^{2-}_{(aq} ---> CaSO4_{(s)}$$

 

[Blueshift5]

Yes, you are correct. The net ionic equation for this reaction would be:

Ca^2+(aq) + 2H^+(aq) + SO_4^2-(aq) --> H_2O(l) + CaSO_4(s)

In order to balance the equation, you would need to add a coefficient of 2 in front of H^+(aq) on the left side, resulting in:

Ca^2+(aq) + 2H^+(aq) + SO_4^2-(aq) --> 2H_2O(l) + CaSO_4(s)

Then, you would separate the aqueous ions on the left side, leaving the solid CaSO_4 unchanged, giving you the final net ionic equation:

Ca^2+(aq) + 2H^+(aq) + SO_4^2-(aq) --> 2H^+(aq) + 2OH^-(aq) + CaSO_4(s)