Ksp for calcium carbonate is 3 x 10-9. if you mix together 100 ml of 0.01 m calcium chloride with 100 ml of 10-5 m solution of potassium carbonate, will a precipitate form?

The Ksp is the solubility product and will tell us whether or not a solid will precipitate. The Ksp for calcium carbonate is shown below:

CaCO₃ (s) → Ca²⁺ + CO₃²⁻     Ksp = 3 x 10⁻⁹ = [Ca²⁺][CO₃²⁻]

We are combining 100 mL of 0.01 M CaCl₂ and 100 mL of 10⁻⁵M K₂CO₃. We can determine the amount of moles of calcium and carbonate ions present and find the new concentrations. The new total volume of the reaction mixture is 200 mL. We can use the formula C1V1 = C2V2

(0.01 M)(100 mL) = C2 (200 mL)
C2 = [Ca²⁺] = 0.005 M

(10⁻⁵ M)(100 mL) = C2 (200 mL)
C2 = [CO₃²⁻] = 5 x 10⁻⁶ M

CaCl₂ + K₂CO₃ → CaCO₃ + 2KCl

We can determine the solubility quotient for CaCO₃ based on the two new concentrations of ions present. If the solubility quotient, Q, is larger than the Ksp then a precipitate will form. If the Q is less than Ksp, then no precipitate forms.

Q = [Ca²⁺][CO₃²⁻]
Q = (0.005)(5 x 10⁻⁶)
Q = 2 x 10⁻⁸

Ksp = 3 x 10⁻⁹

The value of Q > Ksp, therefore, a precipitate will form.

Yes it would be a physical change.