A solution of barium chromate is prepared by dissolving 6.3 x 10-3 g of this yellow solid in 1.00 L of hot water. Will solid barium chromate precipitate upon cooling to 25°C? Proof?
You want to compare Qsp to Ksp (at 25o C.)
Ag2CrO4(s) ==>2Ag^+ + CrO4^=
Ksp = (Ag^+)^2(CrO4^=) = ?? Look it up in your text or notes (at 25 degrees C).
6.3E-3g/molar mass Ag2CrO4 = mols Ag2CrO4 in 1. Lets call this value x molar.
2x then will be Ag^+ and x will be CrO4^=. Plug those values in and multiply them together. Don't forget to square the Ag^+. This will be Qsp See how Ksp compares with Qsp. Ksp CAN'T be exceeded; therefore, if Qsp is larger, a ppt will occur. If Qsp is smaller, no ppt.
Proof?? You can perform the experiment in the lab but be warned. Ksp calculation IS AT EQUILIBRIUM. What happens if you cool the silver chromate to 25 and no ppt occurs but one, by the calculations, SHOULD. What about the possibility of a supersaturated solution? Then what?
To determine if solid barium chromate will precipitate upon cooling to 25°C, we need to compare the value of Qsp (the ion product) to the value of Ksp (the solubility product). If Qsp is larger than Ksp, a precipitation reaction will occur.
First, we need to find the Ksp value for barium chromate at 25°C. You can look up this value in your textbook or notes.
Next, we calculate the number of moles of barium chromate in the given solution. Divide the mass of the solid (6.3 x 10-3 g) by the molar mass of barium chromate to obtain the number of moles of barium chromate dissolved in 1 L of water. Let's call this value "x" moles.
Since the balanced equation for the dissociation of barium chromate is:
BaCrO4(s) ⟶ Ba2+(aq) + CrO4^2-(aq)
We know that for every 1 mole of barium chromate that dissolves, we obtain 1 mole of barium ions (Ba2+) and 1 mole of chromate ions (CrO4^2-). Therefore, we have x moles of barium ions and x moles of chromate ions in the solution.
To find the Qsp, we need to multiply the concentrations of the ions raised to their stoichiometric coefficients. The concentration of barium ions and chromate ions is equal to x molar. Thus, Qsp = (x)^2(x) = x^3.
Now, we compare Qsp to Ksp. If Qsp is larger than Ksp, a precipitation reaction will occur upon cooling.
Keep in mind that the Ksp calculation is done at equilibrium conditions. However, in this case, we are assuming that the system is not yet at equilibrium since we are cooling the solution. It is important to note that if the Qsp is larger than the Ksp value, a precipitate should form according to the calculations. However, there is a possibility of having a supersaturated solution, meaning that it is temporarily stable but can spontaneously precipitate if disturbed.
Ultimately, to confirm if solid barium chromate will precipitate upon cooling to 25°C, it is advisable to perform the experiment in the laboratory and observe the actual outcome.