Question

the ksp for silver chromate (ag2cro4) is 1.1x10-12. using the same initial set up of 10ml of 1m silver nitrate is combined with 25ml of 0.1m sodium chromate, what is the reaction quotient (q) of the silver chromate? (is a precipitate going to form?)

Answer 1

The reaction quotient (Q) for the formation of silver chromate (Ag2CrO4) is calculated to be 0.0714, which is greater than the Ksp value of 1.1 × 10^-12. Therefore, a precipitate of silver chromate will form.

Step-by-step explanation:

To find the reaction quotient (Q) of silver chromate (Ag2CrO4), we need to determine the concentrations of silver ions (Ag+) and chromate ions (CrO4^2-) in the solution. According to the given information, the initial setup includes 10 mL of 1 M silver nitrate (AgNO3) and 25 mL of 0.1 M sodium chromate (Na2CrO4).

To calculate the concentration of Ag+ ions, we divide the number of moles by the total volume in liters: [Ag+] = 0.01 mol / 0.01 L = 1 M. For CrO4^2- ions, we use the same calculation: [CrO4^2-] = 0.0025 mol / 0.035 L ≈ 0.0714 M.

Therefore, the reaction quotient (Q) for the formation of silver chromate (Ag2CrO4) is Q = [Ag+]^2[CrO4^2-] = (1 M)^2(0.0714 M) = 0.0714. Since the Ksp for Ag2CrO4 is 1.1 × 10^-12, Q is greater than Ksp, indicating that a precipitate of silver chromate will form.