How many grams of solid sodium sulfate can be added to 1000L of a 0.02mol/L solution of calcium chloride before a precipitate forms? (Ksp for calcium is 6.1x10^-5)
CaSO4 --> Ca^+2 + SO4^=
Ksp = (Ca^+2)(SO4^=)
Substitute 0.02M for (Ca^+2) and calculate (SO4^=) needed to cause ppt of CaSO4.
That will be in M; therefore, for 1 L of solution that will be the moles.
Then grams Na2SO4 = moles x molar mass Na2SO4.
But why 1L of the solution? Why not use the 1000L?
Because I am so used to working problems with 1 L that I just read 1 L into it. Excuse me please.
You are right. After you know the concn of SO4 needed to ppt CaSO4 (in 1 L), then multiply that by molar mass Na2SO4 to find grams for 1 L, then multiply by 1000 to obtain grams Na2SO4 for 1000 L.
To determine the maximum amount of solid sodium sulfate that can be added before a precipitate forms, we need to consider the solubility product constant (Ksp) and the stoichiometry of the reaction.
First, let's write the balanced equation for the reaction between calcium chloride (CaCl2) and sodium sulfate (Na2SO4):
CaCl2(aq) + Na2SO4(aq) → 2NaCl(aq) + CaSO4(s)
According to the equation, for every 1 mole of calcium chloride, we would need 1 mole of sodium sulfate to produce 1 mole of calcium sulfate (which is the precipitate formed).
Given that the concentration of the calcium chloride solution is 0.02 mol/L and the volume is 1000 L, we can calculate the number of moles of calcium chloride:
Moles of CaCl2 = concentration × volume
Moles of CaCl2 = 0.02 mol/L × 1000 L
Moles of CaCl2 = 20 mol
Since we need an equal number of moles of sodium sulfate, we can calculate the maximum number of moles of sodium sulfate that can react with the calcium chloride:
Moles of Na2SO4 = 20 mol
Now, to convert the number of moles of sodium sulfate to grams, we need to know the molar mass of sodium sulfate. The molar mass of Na2SO4 is approximately 142.04 g/mol.
Mass of Na2SO4 = Moles of Na2SO4 × Molar mass of Na2SO4
Mass of Na2SO4 = 20 mol × 142.04 g/mol
Mass of Na2SO4 = 2840.8 g
Therefore, the maximum amount of solid sodium sulfate that can be added to 1000 L of the 0.02 mol/L calcium chloride solution before a precipitate forms is 2840.8 grams.