How many moles of Calcium chloride can be added to 1.5L of 0.020potassium sulphate before a precipitate is expected? Assume that the volume is not chnaged significantly.
CaSO4 ==> Ca^+2 + SO4^=
Ksp = (Ca^+)(SO4^=) = ??. Look it up.
(SO4^=) = 0.02 M
Use Ksp to calculate (Ca^+2).
The answer will be the concentration in mols/L. Calculate how many mols that will be in 1.5 L.
Post your work if you get stuck.
0.0018 moles of calcium chloride
0.00375 moles
First, let's look up the value of the Ksp for calcium sulfate (CaSO4). The Ksp for calcium sulfate is approximately 4.93 x 10^-5 at 25 degrees Celsius.
Next, we need to calculate the concentration of the sulfate ion (SO4^=) in potassium sulfate (K2SO4) using the given information. The concentration of potassium sulfate is 0.020 M, and since one mole of potassium sulfate produces one mole of sulfate ion, the concentration of sulfate ion is also 0.020 M.
Now we can use the Ksp expression to calculate the concentration of calcium ion (Ca^+2):
Ksp = [Ca^+2][SO4^=]
4.93 x 10^-5 = [Ca^+2][0.020]
Let's solve for [Ca^+2]:
[Ca^+2] = (4.93 x 10^-5) / (0.020)
[Ca^+2] ≈ 2.46 x 10^-3 M
The concentration of calcium ion is approximately 2.46 x 10^-3 M.
To find the number of moles of calcium chloride (CaCl2) that can be added before a precipitate is expected, we need to consider that calcium chloride dissociates into one calcium ion (Ca^+2) and two chloride ions (Cl^-). Since the concentration of calcium ion is 2.46 x 10^-3 M, the concentration of calcium chloride should be twice that to provide an equal concentration of calcium ions.
Therefore, the maximum number of moles of calcium chloride that can be added to 1.5 L of 0.020 M potassium sulfate before a precipitate is expected is:
(2 x 2.46 x 10^-3 M) x 1.5 L = 7.38 x 10^-3 moles
To find the number of moles of calcium chloride that can be added before a precipitate is expected, we need to calculate the concentration of calcium ions (Ca^+2) using the given information and the solubility product constant (Ksp) of calcium sulfate (CaSO4).
1. Start by looking up the Ksp value for calcium sulfate (CaSO4) in a reliable source or reference. Let's assume it is 2.4 x 10^-4 (this is just an example, the actual value may vary).
2. We are given the concentration of sulfate ions (SO4^=) as 0.020 M.
3. Using the formula for Ksp: Ksp = [Ca^+2][SO4^=], we can rearrange it to solve for [Ca^+2]:
[Ca^+2] = Ksp / [SO4^=]
Substituting the values, we have:
[Ca^+2] = (2.4 x 10^-4) / (0.020)
= 1.2 x 10^-2 M
4. Now we know the concentration of calcium ions, but we need to calculate the number of moles of calcium chloride that can be added to 1.5 L of potassium sulfate (K2SO4) solution.
The concentration in moles per liter (mol/L) is the same as the moles per liter (M) concentration. Therefore, the concentration of calcium ions is 1.2 x 10^-2 mol/L.
5. To find the number of moles in 1.5 L of solution, we multiply the concentration by the volume:
Number of moles = concentration (mol/L) * volume (L)
= (1.2 x 10^-2 mol/L) * (1.5 L)
= 1.8 x 10^-2 moles
Therefore, we can add up to 1.8 x 10^-2 moles of calcium chloride to 1.5 L of the potassium sulfate solution before a precipitate is expected, assuming the volume does not significantly change.