I recently did a lab about double replacements and precipitates, my first post lab question asks me to write a proposed reaction for each double replacement reaction I did, in Molecular, Ionic and Net ionic form.
My first reaction was KI and Ca(NO3)2
The issues I am having is that based off of my lab (Where it appeared that no reaction happend) as well as the rules of solubility
1) All Salts of Group 1A are soluble (K)
2) All salts cotaining nitrate (NO3) are soluble
3) All iodides (I) are soluble
This leaves Ca as the only one insoluble.
Because of this I have no idea how to write the equations (especially the net ionic)
This is how I wrote the Molecular:
KI(aq) + Ca(NO3)2 (aq) -> K(NO3)2(aq) + CaI(aq)
And the Ionic
K^+1 (aq) + I^-1(aq) + Ca^+2(aq) + (NO3)2^-1 (aq) -> K^+1 (aq) + NO3(2)^-1 (aq) + Ca^-2(aq) + I -1 (aq)
However I have no idea how to answer the Net Ionic
*Apologizes for the long question, thank you in advance! -Kat
I am unsure about your question but perhaps I can give you a start and you can let me know from there. First, I think the double replacement/precipitation thing idea is that a ppt should form (at least in some cases) for the net ionic equation to be useful. But first let me talk about your solubility rules. I don't buy all of them.
1) All Salts of Group 1A are soluble (K). This is mostly trule although KClO4 has been used to determine K^+
2) All salts containing nitrate (NO3) are soluble OK
3) All iodides (I) are solubleThis definitely is not true. AgI, PbI2, and several others are extremely insoluble. CaI2, however, is soluble.
This leaves Ca as the only one insoluble.Several Ca salts are insoluble such as CaC2O4 but some are soluble. For example CaI2 is soluble.
Because of this I have no idea how to write the equations (especially the net ionic)
Now to your net ionic equation you wrote for Ca(NO3)2 + 2KI ==> CaI2 + 2KNO3. Take a look at your total ionic equation. To get the NET ionic equation you cancel those ions that appear on BOTH side of the equation. I see Ca^2+ on both sides, K^+ on both sides, NO3^- on both sides, and I^- on both sides. So ALL of the ions cancel. Why? Because there is no net reaction. When CaI2 and KNO3 are mixed, there is no reaction. Both salts dissociate, the ions mix and you have a mixture of both salts. There is no ppt that forms and nothing to drive the reaction from left to right. You get a mixing and that's all. If I see a reaction like that I just write no reaction on the product side. I think your confusion is caused somewhat by the erroneous information about the solubility of salts etc. Let me know if this helps with the confusion. I think much of this will become clearer when you have a ppt that forms on the product side. Sorry about the long response.
No need to apologize, Kat! I'm here to help you out.
To write the net ionic equation, we need to know which species are actually involved in the reaction. Based on your information and solubility rules, you correctly identified that Ca(NO3)2 is soluble since it contains nitrate (NO3). On the other hand, KI is also soluble because it contains potassium (Group 1A) and iodide (I). Therefore, there is no precipitation reaction between KI and Ca(NO3)2.
So, the net ionic equation would be empty because there are no ions participating in the reaction. It would be written as:
No reaction
Since there are no precipitates formed, the net ionic equation doesn't have any species to represent. When there is no reaction occurring, the net ionic equation is left empty.
Keep in mind that it's essential to refer to the solubility rules to determine the solubility of compounds and predict double replacement reactions. In this case, the solubility rules helped you recognize that both KI and Ca(NO3)2 are soluble, preventing a reaction from occurring.
If you have any more questions, feel free to ask!