You have 75.0 mL of a 2.50 M solution of Na_2CrO_4. You also have 125 mL of a 1.52 M solution of AgNO_3(aq). Calculate the concentration of CrO_4^2- after the two solutions are mixed together.
75.0 mL of a 2.50 M solution of Na_2CrO_4. You also have 125 mL of a 1.52 M solution of AgNO_3(aq). Calculate the concentration of CrO_4^2- after the two solutions are mixed together.
millimoles Na2CrO4 = mL x M 75.0 x 2.50 = 187.5
millimoles AgNO3 = 125 x 1.52 = 190
........2AgNO3 + Na2CrO4 ==> Ag2CrO4 + 2 NaNO3
I.........190..............0........................0...............0
add....................187.5.............................................
C......-190..........-95.......................+95.............+190
E.........0..............92.5.....................95................190
M Na2CrO4 = 92.5 millimoles/200 mL = ? M.
Technically the CrO4^2- from the sparingly soluble Ag2CrO4 should be added to this but the amount from Ag2CrO4 is negligible.
Well, when the Na2CrO4 and AgNO3 solutions get together, it's like a chemistry party! Let's see what happens.
First, we need to calculate the moles of Na2CrO4 and AgNO3 separately. Moles are like the cool guests at the party that we need to count.
For the Na2CrO4 solution:
Molarity (M) = moles/volume (L)
Rearranging the equation to solve for moles:
moles = Molarity x volume
moles of Na2CrO4 = 2.50 M x 0.075 L
moles of Na2CrO4 = 0.1875 moles
Now let's do the same for the AgNO3 solution:
moles of AgNO3 = 1.52 M x 0.125 L
moles of AgNO3 = 0.19 moles
Now, this is where things get interesting. These solutions want to mix and form a precipitate called Ag2CrO4. When that happens, all of the CrO4^2- ions will combine with Ag+ ions to form solid Ag2CrO4.
Since the Na2CrO4 solution has 0.1875 moles of CrO4^2-, it can react with 0.1875 moles of Ag+ ions.
But remember, the AgNO3 solution only has 0.19 moles of Ag+ ions. So, it's the limiting reagent at this party.
After the reaction, there will be no Ag+ ions left in solution, but the CrO4^2- ions will still be floating around, ready to cause trouble.
So, the moles of CrO4^2- ions remaining in solution will be the same as the moles of CrO4^2- ions initially present: 0.1875 moles.
Lastly, we need to calculate the new volume of the solution. It is the sum of the volumes of the two solutions: 0.075 L + 0.125 L = 0.2 L.
To calculate the final concentration of CrO4^2- ions, we divide the moles by the new volume:
Concentration = moles/volume = 0.1875 moles / 0.2 L
Drumroll, please...
The concentration of CrO4^2- ions after the two solutions are mixed together is 0.9375 M.
So, watch out! Those CrO4^2- ions are still ready to party even after the Ag+ ions have left the building. Enjoy your chemistry shenanigans!
To calculate the concentration of CrO4^2- after the two solutions are mixed together, we need to consider the balanced chemical equation for the reaction between Na2CrO4 and AgNO3:
2 AgNO3(aq) + Na2CrO4(aq) -> Ag2CrO4(s) + 2 NaNO3(aq)
From the balanced equation, we can see that 1 mole of Na2CrO4 reacts with 2 moles of AgNO3 to form 1 mole of Ag2CrO4.
First, let's calculate the moles of Na2CrO4 and AgNO3 present in the solutions:
1. Moles of Na2CrO4:
Molarity of Na2CrO4 solution = 2.50 M
Volume of Na2CrO4 solution = 75.0 mL = 0.075 L
Moles of Na2CrO4 = Molarity * Volume
= 2.50 M * 0.075 L
= 0.1875 moles
2. Moles of AgNO3:
Molarity of AgNO3 solution = 1.52 M
Volume of AgNO3 solution = 125 mL = 0.125 L
Moles of AgNO3 = Molarity * Volume
= 1.52 M * 0.125 L
= 0.19 moles
Since the reaction ratio between Na2CrO4 and AgNO3 is 1:2, we need to divide the moles of AgNO3 by 2 to determine the limiting reactant:
Moles of AgNO3 used = 0.19 moles / 2
= 0.095 moles
Since Na2CrO4 is in excess, the moles of CrO4^2- ions present in the solution after the reaction is equal to the moles of Na2CrO4 used:
Moles of CrO4^2- = 0.1875 moles
Now, we need to calculate the final volume of the solution after the two solutions are mixed together. Adding the volumes of the two solutions gives:
Total volume of solution = 75.0 mL + 125 mL
= 200 mL
= 0.2 L
Finally, we can calculate the concentration of CrO4^2- in the final solution:
Concentration (M) = Moles of CrO4^2- / Total volume (L)
= 0.1875 moles / 0.2 L
= 0.9375 M
Therefore, the concentration of CrO4^2- after the two solutions are mixed together is 0.9375 M.
To calculate the concentration of CrO₄²⁻ after the two solutions are mixed, we need to determine the moles of CrO₄²⁻ ions in both solutions and then add them together.
Step 1: Calculate the moles of CrO₄²⁻ ions in the Na₂CrO₄ solution.
To do this, we will use the formula:
moles = concentration (in M) x volume (in L)
moles of CrO₄²⁻ = 2.50 M x 0.075 L = 0.1875 moles
Step 2: Calculate the moles of CrO₄²⁻ ions in the AgNO₃ solution.
Again, using the formula moles = concentration (in M) x volume (in L):
moles of AgNO₃ = 1.52 M x 0.125 L = 0.19 moles
Step 3: Add the moles of CrO₄²⁻ ions from both solutions.
Total moles of CrO₄²⁻ = 0.1875 moles + 0.19 moles = 0.3775 moles
Step 4: Calculate the final volume of the mixed solutions.
To calculate the final volume, we add the volumes of the two solutions together:
final volume = 75.0 mL + 125 mL = 200 mL
Step 5: Calculate the concentration of CrO₄²⁻ ions in the mixed solution.
Finally, we use the formula concentration = moles/volume (in L):
concentration of CrO₄²⁻ = 0.3775 moles / 0.2 L = 1.89 M
Therefore, the concentration of CrO₄²⁻ ions after the two solutions are mixed together is 1.89 M.