Question

Potassium chromate is slowly added to a solution containing 0.20 M AgNO3 and 0.20 M Ba(NO3)2. Describe what happens if the Ksp for Ag2CrO4 is 1.1 × 10-12 and the Ksp of BaCrO4 is 1.2 × 10-10.

Answer 1

Answer:

Ag2CrO4 will first start precipitating when $[CrO_4^{-2}] or [K_2CrO_4]=2.75 \times10^{-11} M$

BaCrO4 will start precipitating next when $[CrO_4^{-2}] or [K_2CrO_4]=6.0\times10^{-10} M$

Explanation:

When potassium chromate (K2CrO4) is added slowly to the solution containing AgNO3 and Ba(NO3)2, then Ag2CrO4 and BaCrO4 will precipitate.

Ksp for Ag2CrO4 = $1.1\times10^{-12}$

Ksp for BaCrO4 = $1.2\times10^{-10}$

Since, Ksp of AgCrO4 is less as compared to Ksp of BaCrO4, therefore, AgCrO4 will precipitate first.

$Ag_2CrO_4 <=>2Ag^{+} +CrO_4^{-2}$

[Ag]+ = 0.20 M

$Ksp = [Ag^{+}]^{2}[CrO_4^{-2}]=1.1\times10^{-12}$

$(0.20)^{2} \times [CrO_4^{-2}]=1.1\times10^{-12}$

$[CrO_4^{-2}]=2.75 \times10^{-11} M$

Ag2CrO4 will start precipitating when $[CrO_4^{-2}] or [K_2CrO_4]=2.75 \times10^{-11} M$

$BaCrO_4 <=>Ba^{2+}+CrO_4^{-2}$

[Ba]2+ = 0.20 M

$Ksp = [Ba^{2+}][CrO_4^{-2}]=1.2\times10^{-10}$

$(0.20) \times [CrO_4^{-2}]=1.2\times10^{-10}$

$[CrO_4^{-2}]=6.0 \times10^{-10} M$

BaCrO4 will start precipitating when $[CrO_4^{-2}] or [K_2CrO_4]=6.0\times10^{-10} M$

Answer 2

The addition of Potassium chromate results in the first precipitation of silver chromate, and when the concentration of Potassium chromate is reduced to $\rm 6.0\;\times\;10^{-10}$ M, barium chromate starts to precipitate.

What is Ksp?

The Ksp has been the solubility constant for the compound. Increase in the concentration from Ksp results in the precipitation.

The Ksp for silver chromate is $1.1\;\times\;10^{-12}$. The concentration of chromate that initiates the precipitation of silver chromate is:

$\rm Ksp=[Ag^2^+][Cr_2O_4^{2-}]$

The concentration of silver ion from silver nitrate is 0.2 M.

$\rm 1.1\;\times\;10^{-12}=(0.2)^2\;[Cr_2O_4^{2-}]\\\\C_2O_4^{2-}=2.75\;\times\;10^{-11}$

The minimum concentration of chromate that initiates the precipitation of silver chromate is $\rm 2.75\;\times\;10^{-11}$ M.

The concentration of chromate for the precipitation of barium chromate with concentration of barium as 0.2 M, and the Ksp as $1.2\;\times\;10^{-10}$.

$\rm 1.2\;\times\;10^{-10}=0.2\;\times\;[CrO_4^2^-]\\CrO_4^2^-=6.0\;\times\;10^-^1^0$

The concentration of chromate when reduced to $6.0\;\times\;10^{-10}$ M, the precipitation of barium chromate is initiated.

Learn more about Ksp, here:

https://brainly.com/question/4736767