Calculating Yield and Concentrations in a Precipitation Reaction

  • Thread starter alingy1
  • Start date
In summary, a 0.410g sample of sodium fluoride is reacted with 50.00mL of 0.500mol/L calcium nitrate solution, resulting in 0.225g of precipitate being recovered after filtration. The net ionic equation for this reaction is not specified. The maximum mass of precipitate that could be formed is 0.255g (theoretical yield). The percentage yield for this reaction is not given. There is confusion regarding the concentration of the three major ions in the final solution, with one source stating 0.268 M Ca2+ and the textbook answer stating 0.294M. The calculation method used in both cases is not specified. However, it is mentioned that the
  • #1
alingy1
325
0
17. (20 marks) A 0.410g sample of sodium fluoride is dissolved in 25.00mL of water and is then reacted with 50.00mL of 0.500mol/L calcium nitrate solution. After the mixture is filtered, 0.225g of precipitate is recovered.
a ) Write the net ionic equation for the reaction which occurs.
b) Calculate the maximum mass of precipitate that could be formed (theoretical
yield).
c) Calculate the percentage yield for this reaction.
d) Calculate the concentrations of the three major ions in the final solution.


https://ca.answers.yahoo.com/question/index?qid=20111210111858AA1ObOY
For d) This website says :Mol Ca2+ remaining in solution: 0.025 mol Ca2+ initially - 4.88X10^-3 mol CaF2 formed = 0.0201 mol Ca2+ remaining / 0.075 L = 0.268 M Ca2+

My textbook answer says the same thing.

BUT WHY! That is the result of theoretical maximum yield 100%... I get 0.294M using 0.225g.
Can you back me up?
 
Physics news on Phys.org
  • #2
I feel dumb :S It looks really simple but I don't get that 0.268M! Why are they talking about full dissociation?!
 
  • #3
It is a rather lousy question IMHO. I have no idea what is the "correct" answer here. That is, concentrations can be calculated from the solubility product, but that's an entirely different problem then.
 
  • #4
I think they waant to us to keep up with the 0.255g.. But apparently we are supposed to assume full dissociation!
 
  • #5
I agree. Bad formulation. And even their answer does not make sense since they ask for 0.225g dissociation.
 

Related to Calculating Yield and Concentrations in a Precipitation Reaction

1. What is dissolution stochiometry?

Dissolution stochiometry is a branch of chemistry that deals with the quantitative relationship between the amount of a substance dissolved and the amount of solvent used. It involves calculations and measurements to determine the proportions of substances involved in a dissolution reaction.

2. How is dissolution stochiometry different from regular stochiometry?

Dissolution stochiometry focuses specifically on the dissolution of substances in a solvent, whereas regular stochiometry deals with the quantitative relationships in chemical reactions in general. In dissolution stochiometry, the solvent is considered as part of the reaction and its quantity is taken into account in calculations.

3. What are the key factors that affect dissolution stochiometry?

The key factors that affect dissolution stochiometry include the nature of the solute and solvent, temperature, pressure, and agitation. These factors can influence the rate and extent of dissolution, and therefore have an impact on the stochiometric calculations.

4. How is dissolution stochiometry used in practical applications?

Dissolution stochiometry is used in various industries, such as pharmaceuticals, food and beverage, and environmental science. It is used to determine the solubility of substances, which is important in drug development, food production, and pollution control. It is also used in quality control to ensure the proper concentration of substances in a solution.

5. Can dissolution stochiometry be applied to reactions other than dissolution?

Yes, the principles of dissolution stochiometry can be applied to other types of reactions, such as precipitation and acid-base neutralization. In these cases, the solvent is still considered as part of the reaction and its quantity is taken into account in the calculations. However, the specific equations and variables used may differ from those in dissolution stochiometry.

Similar threads

  • Biology and Chemistry Homework Help
Replies
2
Views
2K
  • Biology and Chemistry Homework Help
Replies
1
Views
3K
  • Biology and Chemistry Homework Help
Replies
4
Views
2K
Replies
1
Views
2K
  • Biology and Chemistry Homework Help
Replies
7
Views
2K
  • Biology and Chemistry Homework Help
Replies
5
Views
2K
  • Biology and Chemistry Homework Help
Replies
5
Views
2K
  • Biology and Chemistry Homework Help
Replies
1
Views
3K
Replies
4
Views
3K
  • Biology and Chemistry Homework Help
Replies
2
Views
3K
Back
Top