For CaCrO4, Ksp= 7.1x 10^-4. How many grams of Na2CrO4 must be added to 200 mL of .250 M Ca(NO3)2 to just initiate precipitation?
.......CaCrO4 ==> Ca^2+ + CrO4^2-
I......solid.......0.......0
C......solid.......x.......x
E......solid.......x.......x
........Ca(NO3)2 ==> Ca^2+ + NO3^2-
I........0.25M........0.......0
C.......-0.25.......0.25......0.25
E..........0........0.25......0.25
Note that CaCrO4 is not soluble and uses Ksp. Ca(NO3)2 is soluble so all of it has dissolved and there is no Ca(NO3)2 remaining.
Total (Ca^2+) = x from CaCrO4 and 0.25 from Ca(NO3)2
Ksp = (Ca^2+)(CrO4^2-)
You know Ksp, you know (Ca^2+) [from above it is x + 0.25), calculate (CrO4^2-) which will have units of mols/L. Also that will be (Na2CrO4) in mols/L. Convert to grams/L [g = mols x molar mass) and convert that to grams in 200 mL.
Post your work if you get stuck.
To calculate the amount of Na2CrO4 needed to just initiate precipitation, we need to use the concept of the solubility product constant (Ksp) and stoichiometry.
The balanced equation for the precipitation reaction is:
Ca(NO3)2 + Na2CrO4 → CaCrO4 + 2NaNO3
From the balanced equation, it can be seen that one mole of Ca(NO3)2 reacts with one mole of Na2CrO4 to form one mole of CaCrO4. Therefore, the stoichiometric ratio between Ca(NO3)2 and Na2CrO4 is 1:1.
First, we need to calculate the moles of Ca(NO3)2 in 200 mL of 0.250 M solution:
Moles of Ca(NO3)2 = concentration (M) × volume (L)
Moles of Ca(NO3)2 = 0.250 M × 0.200 L
Moles of Ca(NO3)2 = 0.05 moles
Since the stoichiometric ratio between Ca(NO3)2 and Na2CrO4 is 1:1, the moles of Na2CrO4 needed are also 0.05 moles.
Now, we need to convert the moles of Na2CrO4 to grams. To do this, we will use the molar mass of Na2CrO4.
The molar mass of Na2CrO4 = (2 × atomic mass of Na) + atomic mass of Cr + (4 × atomic mass of O)
The atomic masses can be found on the periodic table.
Molar mass of Na2CrO4 = (2 × 22.99 g/mol) + 51.996 g/mol + (4 × 16.00 g/mol)
Molar mass of Na2CrO4 = 105.988 g/mol
Now, we can calculate the grams of Na2CrO4 needed:
Grams of Na2CrO4 = moles of Na2CrO4 × molar mass of Na2CrO4
Grams of Na2CrO4 = 0.05 moles × 105.988 g/mol
Grams of Na2CrO4 = 5.2994 grams
Therefore, approximately 5.30 grams of Na2CrO4 must be added to 200 mL of 0.250 M Ca(NO3)2 to just initiate precipitation.
To determine how many grams of Na2CrO4 must be added to initiate precipitation, we need to calculate the maximum amount of CaCrO4 that can dissolve in the given solution of Ca(NO3)2. This can be achieved using the concept of solubility product constant, Ksp.
The balanced chemical equation for the precipitation reaction is:
Ca(NO3)2 + Na2CrO4 → CaCrO4 + 2NaNO3
The molar solubility of CaCrO4 is equal to the concentration of dissolved CaCrO4, assuming it reaches equilibrium. Let's assume the molar solubility of CaCrO4 is 'x' mol/L.
The equilibrium expression is expressed as:
Ksp = [CaCrO4] * [NaNO3]^2
Since we are assuming the molar solubility of CaCrO4 is 'x', we can write:
Ksp = x * (0.250 M)^2
Now, substitute the given value for Ksp into the equation:
7.1 × 10^-4 = x * (0.250 M)^2
Solving for x:
x = (7.1 × 10^-4) / (0.250 M)^2
x ≈ 1.136 × 10^-2 M
Since the concentration of Na2CrO4 should be twice the concentration of CaCrO4 (according to the balanced equation), the concentration of Na2CrO4 is approximately 2 * 1.136 × 10^-2 M.
Now, we can calculate the number of moles of Na2CrO4 using the given volume of 200 mL (0.200 L) and the concentration:
moles = concentration * volume
moles of Na2CrO4 = (2 * 1.136 × 10^-2 M) * 0.200 L
Finally, to find the grams of Na2CrO4, we need to multiply the moles by the molar mass of Na2CrO4.
You can calculate the molar mass of Na2CrO4 by adding up the atomic masses:
Na: 22.99 g/mol (2 atoms)
Cr: 52.00 g/mol
O: 16.00 g/mol (4 atoms)
Molar mass of Na2CrO4 = 22.99 g/mol * 2 + 52.00 g/mol + 16.00 g/mol * 4
Now, multiply the moles by the molar mass to find the grams:
grams of Na2CrO4 = moles of Na2CrO4 * molar mass of Na2CrO4
After performing these calculations, you will determine the grams of Na2CrO4 required to just initiate precipitation in the given solution.