0.10 M potassium chromate is slowly added to a solution containing 0.50 M AgNO3 and 0.50 M Ba(NO3)2. What is the Ag+ concentration when BaCrO4 just starts to precipitate? The Ksp for Ag2CrO4 and BaCrO4 are 1.1 x 10-12 and 1.2x10^-10 respectively
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To determine the Ag+ concentration when BaCrO4 just starts to precipitate, we need to calculate the concentrations of Ag+ and CrO4^2- ions at equilibrium.
First, let's set up the balanced chemical equation for the precipitation reaction:
Ba(NO3)2 + K2CrO4 → BaCrO4 + 2 KNO3 (1)
According to the stoichiometry of the reaction, 1 mole of BaCrO4 is formed for each mole of Ba(NO3)2 reacted, and 2 moles of K2CrO4 are required for every mole of BaCrO4 formed.
Next, we need to calculate the initial concentrations of Ag+ and CrO4^2- ions in solution. Since both AgNO3 and Ba(NO3)2 are 0.50 M, their initial concentrations are known.
Now, let's calculate the Ag+ concentration when BaCrO4 just starts to precipitate.
1. Write the solubility product expression for Ag2CrO4:
Ksp = [Ag+]^2[CrO4^2-]
2. Determine the concentration of CrO4^2- ions at equilibrium:
Since 1 mole of BaCrO4 is formed, we need to account for the 2 moles of K2CrO4 reacted.
The initial concentration of K2CrO4 is 0.10 M, and since the reaction goes to completion (limiting reagent), all of the K2CrO4 will react.
Therefore, the concentration of CrO4^2- ions is 2 * 0.10 M = 0.20 M.
3. Use the solubility product constant (Ksp) for Ag2CrO4 to calculate the Ag+ concentration when BaCrO4 just starts to precipitate:
Ksp = [Ag+]^2[CrO4^2-]
Rearranging the equation, [Ag+] = sqrt(Ksp/[CrO4^2-])
Plugging in the values, [Ag+] = sqrt(1.1x10^-12 / 0.20) = 5.24x10^-7 M
Therefore, the Ag+ concentration when BaCrO4 just starts to precipitate is approximately 5.24x10^-7 M.