1. We dissolve 0.005 mole of strontium chromate in 1 L of solution at 25°C. Will a precipitate be formed?
Assuming that the solution is in water...
Strontium chromate is insoluble in water so yes a precipitate will form.
To determine if a precipitate will be formed when 0.005 mole of strontium chromate is dissolved in 1 L of solution at 25°C, you need to calculate the solubility product constant (Ksp) of strontium chromate and compare it to the ion product (Q).
The balanced chemical equation for the dissociation of strontium chromate (SrCrO4) is:
SrCrO4(s) ↔ Sr2+(aq) + CrO4^2-(aq)
The expression for the Ksp of strontium chromate is:
Ksp = [Sr2+][CrO4^2-]
To determine the value of Ksp, you will need to refer to a reliable source or use experimental data. For the purpose of this explanation, let's assume that the Ksp of strontium chromate is 2.5 x 10^-5.
Now, calculate the ion product (Q) by substituting the concentrations of the ions into the Ksp expression:
Q = [Sr2+][CrO4^2-]
Since you dissolved 0.005 mole of strontium chromate in 1 L of solution, the concentration of Sr2+ and CrO4^2- ions will be equal to 0.005 M.
Substituting the values into the expression, we get:
Q = (0.005)(0.005) = 2.5 x 10^-5
If Q is less than Ksp (Q < Ksp), then the system is unsaturated and no precipitate will form. However, if Q is equal to or greater than Ksp (Q ≥ Ksp), then the system is saturated, and a precipitate will form.
In this case, Q (2.5 x 10^-5) is equal to the assumed Ksp value (2.5 x 10^-5). Therefore, the system is saturated, and a precipitate will form.