Solutions of silver nitrate and sodium chromate are mixed. What is the net ionic equation?
I got: 2Ag(aq) + CrO4-2(aq) -> Ag2CrO4(s)
is that right?
I would add a +sign for charge on the Ag ion.
Yes, a + sign should b added to the Ag ion
Yes, your net ionic equation is correct. When silver nitrate and sodium chromate are mixed, a double replacement reaction takes place. The silver cation (Ag+) from silver nitrate reacts with the chromate anion (CrO4^2-) from sodium chromate to form solid silver chromate (Ag2CrO4). The net ionic equation for this reaction can be written as:
2Ag+(aq) + CrO4^2-(aq) → Ag2CrO4(s)
To determine the net ionic equation for the given reaction between silver nitrate (AgNO3) and sodium chromate (Na2CrO4), we need to first write the balanced chemical equation.
The balanced equation for the reaction is as follows:
2AgNO3(aq) + Na2CrO4(aq) → Ag2CrO4(s) + 2NaNO3(aq)
Now, let's write the ionic equation by representing the reactants and products as their respective ions:
2Ag+(aq) + 2NO3-(aq) + 2Na+(aq) + CrO4^2-(aq) → Ag2CrO4(s) + 2Na+(aq) + 2NO3-(aq)
Next, we need to cancel out the spectator ions that appear on both sides of the equation (ions that do not participate in the reaction). In this case, the spectator ions are Na+ and NO3-.
After canceling out the spectator ions, the net ionic equation is:
2Ag+(aq) + CrO4^2-(aq) → Ag2CrO4(s)
So, your initial answer is correct!
The net ionic equation for the reaction between silver nitrate and sodium chromate is:
2Ag(aq) + CrO4-2(aq) -> Ag2CrO4(s)