30.0 mL of 0.10 M Ca(NO3)2 and 15.0 mL of 0.20 M Na3PO4 solutions are mixed. After the reaction is complete, which of these ions has the lowest concentration in the final solution?
A) Na+ B) NO3- C) Ca+2 D) PO4-3
Answer is C.. How do you solve it though?!
If you imagine the process occuring in 2 discrete steps, mixing and precipitation, you get the following.
The mixing step gives you:
30.0 mL of 0.10 M Ca(NO3)2 = 3 mmol Ca(NO3)2 = 3 mmol Ca(2+) + 6 mmol NO3(-).
15.0 mL of 0.20 M Na3PO4 = 3 mmol Na3PO4 = 9 mmol Na(+) + 3 mmol PO4(3-)
Then the precipitation step is
3 Ca(2+) + 2 PO4(3-) ---> Ca3(PO4)2 which forms a solid.
Note that you have a 1:1 ratio in solution but a 3:2 ratio in the reaction. Ca(2+) is the limiting reagent. So it will be used up in the precipitation reaction faster than PO4(3-).
This means that Ca(2+) will have the lowest concentration in the end.
Oh, chemistry, always calculating things and making life complicated! But don't worry, I have a joke for you while we figure this out. Why did the chemist wear two layers of underwear? Because he wanted to have a double bond!
Now, let's get back to solving this. To figure out which ion has the lowest concentration, we need to look at the balanced chemical equation. The equation for the reaction between calcium nitrate (Ca(NO3)2) and sodium phosphate (Na3PO4) is:
3Ca(NO3)2 + 2Na3PO4 → Ca3(PO4)2 + 6NaNO3
From the balanced equation, we can see that for every 3 moles of Ca(NO3)2, we get 1 mole of Ca3(PO4)2. This means that the concentration of Ca+2 ions in the final solution will be lower than the concentrations of Na+ and NO3- ions.
So, the answer is C) Ca+2. That poor calcium ion just couldn't keep up with the sodium and nitrate ions hanging out in the solution!
To solve this problem, you need to determine the concentration of each ion in the final solution after the reaction between Ca(NO3)2 and Na3PO4 is complete.
Step 1: Calculate the number of moles of each ion in the solution.
The number of moles (n) can be calculated by using the formula:
n = C x V
Where C is the concentration of the solution and V is the volume of the solution.
For Ca2+ ion:
n(Ca2+) = C(Ca2+) x V(Ca(NO3)2)
= 0.10 M x 30.0 mL
= 0.003 mol
For PO43- ion:
n(PO43-) = C(PO43-) x V(Na3PO4)
= 0.20 M x 15.0 mL
= 0.003 mol
Step 2: Determine the moles of ions reacted based on their stoichiometry.
The balanced chemical equation for the reaction between Ca(NO3)2 and Na3PO4 is:
3Ca(NO3)2 + 2Na3PO4 → Ca3(PO4)2 + 6NaNO3
From the balanced equation, we can see that for every 3 moles of Ca(NO3)2 reacted, we will have 1 mole of Ca3(PO4)2 formed. Similarly, for every 2 moles of Na3PO4 reacted, we will have 1 mole of Ca3(PO4)2 formed.
Since we have equal moles of Ca2+ and PO43-, it means that all the moles of the ions have reacted completely to form Ca3(PO4)2. Therefore, the concentration of Ca2+ ion would be zero in the final solution.
So, the correct answer is C) Ca+2 has the lowest concentration in the final solution.
To solve this problem, we need to determine the products of the reaction between Ca(NO3)2 and Na3PO4, and then compare the concentrations of the ions in the final solution.
The balanced chemical equation for the reaction between Ca(NO3)2 and Na3PO4 can be written as:
3Ca(NO3)2 + 2Na3PO4 -> Ca3(PO4)2 + 6NaNO3
From the equation, we can see that 3 moles of Ca2+ (from Ca(NO3)2) react with 2 moles of PO43- (from Na3PO4) to form 1 mole of Ca3(PO4)2. The remaining ions, 6 moles of Na+ (from Na3PO4) and 6 moles of NO3- (from Ca(NO3)2), remain in the solution.
To calculate the concentration of each ion in the final solution, we need to first find the number of moles of each ion.
1. For Ca2+:
The initial volume of Ca(NO3)2 solution is 30.0 mL, which is equal to 30.0/1000 = 0.0300 L. Therefore, the number of moles of Ca2+ is:
moles of Ca2+ = volume (L) x concentration (M)
moles of Ca2+ = 0.0300 L x 0.10 M = 0.00300 moles
2. For Na+:
The initial volume of Na3PO4 solution is 15.0 mL, which is equal to 15.0/1000 = 0.0150 L. Therefore, the number of moles of Na+ is:
moles of Na+ = volume (L) x concentration (M)
moles of Na+ = 0.0150 L x 0.20 M = 0.00300 moles
3. For NO3-:
The initial volume of Ca(NO3)2 solution is 30.0 mL, which is equal to 30.0/1000 = 0.0300 L. Therefore, the number of moles of NO3- is:
moles of NO3- = volume (L) x concentration (M)
moles of NO3- = 0.0300 L x 0.10 M = 0.00300 moles
4. For PO43-:
The initial volume of Na3PO4 solution is 15.0 mL, which is equal to 15.0/1000 = 0.0150 L. Therefore, the number of moles of PO43- is:
moles of PO43- = volume (L) x concentration (M)
moles of PO43- = 0.0150 L x 0.20 M = 0.00300 moles
From the calculations above, we can see that the number of moles of all ions is the same, which means they have the same concentration in the final solution. Therefore, answer C) Ca+2 is incorrect.
The correct answer is A) Na+, which has the lowest concentration in the final solution.