Yr11 Chem Q (2 Viewers)

[Axio]

Hey could someone show me how to do this question? After looking at the answers I think I understand how to do it now but I'm not sure about something they did. Thanks .

Equal volumes of lead II nitrate solution and sodium iodide solution, each with a concentration of 0.10mol/L are mixed. A yellow precipitate of lead II iodide results. Calculate the concentration of each ion remaining in solution.

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I just thought that in 0.05mol of Pb(NO3)2 there was 0.10mol of NO3-, which would make it 0.2mol/L NO3-. Is that wrong?

 

[zhertec]

Equal volumes of lead II nitrate solution and sodium iodide solution, each with a concentration of 0.10mol/L are mixed. A yellow precipitate of lead II iodide results. Calculate the concentration of each ion remaining in solution.

NOTE:When I write M after a numerical value, it stands for Moles/L, so concentration form.
1. Write out the balanced equation: Pb(NO3)2 (aq) + 2NaI (aq) --> PbI2 (s) + 2NaNO3 (aq)
2. Plug in all the given info: 0.1M 0.1M
3. Most important step (probably), find out which of the substances is the limiting reagent i.e. which one will be used up first in the reaction. This is important as working with precipitates, the amount of precipitates formed is proportional to the limiting reagent.
4. So basically for Pb(NO3)2, it will require 0.2 M of NaI to be fully used up, but there is only 0.1M hence NaI is the limiting reagent.
5. Now only use the Moles/L from the 2NaI to find out the moles of the products.
6. 0.1M of 2NaI will produce 0.05M of PbI2 and 0.1M of 2NaNO3 (using molar ratios etc).
7. Now since all the iodine is used up in the reaction, precipitating with lead, it is no longer an ion (charged) hence it equals 0. Since Sodium is a spectator ion, it will stay in aqueous form, hence stay as 0.1M.
8. Now for the Nitrate (NO3). Since only part of the Lead Nitrate reacted to form the products, there is still an unreacted amount of Nitrate left in the solution, which is 0.05M (using molar ratios from the 2NaI). Hence 0.5M of unreacted lead ions and nitrate ions. The part of the lead nitrate that did react with the solution produced 0.1M of Sodium Nitrate (and since nitrates are also spectators, they stay in aqueous as well), as a result of NO3 staying in ion form, you add the the values of NO3 that takes part in the reaction and the amount that does not take part in the reaction, giving a total amount of 0.15M of NO3 ions.

Hope I helped, if there is still any issues feel free to ask again :3 And if I get anything wrong, someone else correct me xD

 

[Axio]

NOTE:When I write M after a numerical value, it stands for Moles/L, so concentration form.
1. Write out the balanced equation: Pb(NO3)2 (aq) + 2NaI (aq) --> PbI2 (s) + 2NaNO3 (aq)
2. Plug in all the given info: 0.1M 0.1M
3. Most important step (probably), find out which of the substances is the limiting reagent i.e. which one will be used up first in the reaction. This is important as working with precipitates, the amount of precipitates formed is proportional to the limiting reagent.
4. So basically for Pb(NO3)2, it will require 0.2 M of NaI to be fully used up, but there is only 0.1M hence NaI is the limiting reagent.
5. Now only use the Moles/L from the 2NaI to find out the moles of the products.
6. 0.1M of 2NaI will produce 0.05M of PbI2 and 0.1M of 2NaNO3 (using molar ratios etc).
7. Now since all the iodine is used up in the reaction, precipitating with lead, it is no longer an ion (charged) hence it equals 0. Since Sodium is a spectator ion, it will stay in aqueous form, hence stay as 0.1M.
8. Now for the Nitrate (NO3). Since only part of the Lead Nitrate reacted to form the products, there is still an unreacted amount of Nitrate left in the solution, which is 0.05M (using molar ratios from the 2NaI). Hence 0.5M of unreacted lead ions and nitrate ions. The part of the lead nitrate that did react with the solution produced 0.1M of Sodium Nitrate (and since nitrates are also spectators, they stay in aqueous as well), as a result of NO3 staying in ion form, you add the the values of NO3 that takes part in the reaction and the amount that does not take part in the reaction, giving a total amount of 0.15M of NO3 ions.

Hope I helped, if there is still any issues feel free to ask again :3 And if I get anything wrong, someone else correct me xD

Thanks very much! So clearly this isn't the case then:

I just thought that in 0.05mol of Pb(NO3)2 there was 0.10mol of NO3-, which would make it 0.2mol/L NO3-. Is that wrong?

 

[zhertec]

For determining molar ratios etc, from past experiences, I've learnt that you rarely go into the components of molecules and its ratios, unless it asks for something in relation to the molecular weight or the percentage of a certain molecule i.e. so you rarely if ever, need to think that Pb(NO3)2 can split up into its specific ratios of Pb and NO3. Well you can but it'd be a lot more complicated. So basically when working with moles and concentration, mostly worry about the numbers in the front of the molecule.