No, you don't ignore sulfate. The sulfate ends up as part of the ionic equation and nitrate cancels in the process of writing the net ionic equation.
1. Write the molecular equation and balance it.
Na2SO4(aq) + Ba(NO3)2(aq) ==>BaSO4(s) + 2NaNO3(aq)
2. Now turn this into an ionic equation. Those materials that are solid are written as solids and not ionized; those that are aq are shows as ions. Gases, if any, stay as gases. Pure liquids (such as water) are not ionized.
2Na^+(aq) + SO4^-2(aq) + Ba^+2(aq) + 2NO3^-(aq) ==> BaSO4(aq) + 2Na^+(aq) + 2NO3^-)(aq)
3. Now we cancel any ions common to both sides. That will be Na^+ and NO3^-. These are spectator ions.
4. What is left is the net ionic equation.
Ba^+2(aq) + SO4^-2(aq) ==> BaSO4(aq)
With some practice, you will be able to dispense with steps 1, 2, and 3, and write the net ionic equation from scratch.'
Ba^+2(aq) + SO4^-2(aq) ==> BaSO4(s)
Could you please tell me how to work out the net balanced equation of 'aqueous solutions of sodium sulfate and barium nitrate when they are mixed?'
*Do we just ignore sulfate and nitrate???????
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