The Question is from neet-chemistry chemical-equilibrium , and it was created by the author Mr. S M HUSSAIN

Question

At 25°C, the solubility product of Ag$_2$CrO$_4$ is 1.1 x 10$^{-12}$. What is the concentration of Ag$^+$ ions in a saturated solution of Ag$_2$CrO$_4$ in 0.1 M K$_2$CrO$_4$?

Mr. S M HUSSAIN · Accepted Answer

## CORE INFORMATION
✓ **Answer**: B
## CONCEPT FRAMEWORK
 **Primary Concept**: Solubility product and common ion effect
 **Related Topics**: Equilibrium, ionic compounds, solubility
 **Prerequisites**: Understanding of solubility product ($K_{sp}$), equilibrium expressions, and the common ion effect.
## SOLUTION PATHWAY
**Given Information**:
  $K_{sp} (Ag_2CrO_4) = 1.1 	imes 10^{-12}$
  $[K_2CrO_4] = 0.1 M$
  $[CrO_4^{2-}] = 0.1 M (due to complete dissociation of K$_2$CrO$_4$)
**Method & Steps**:
 1. Write the equilibrium expression for the dissolution of Ag$_2$CrO$_4$.
  $Ag_2CrO_4(s) ightleftharpoons 2Ag^+(aq) + CrO_4^{2-}(aq)$
  $K_{sp} = [Ag^+]^2[CrO_4^{2-}]$
 2. Use the given $K_{sp}$ and the concentration of chromate ions from K$_2$CrO$_4$ to find the concentration of Ag$^+$ ions.
  $1.1 	imes 10^{-12} = [Ag^+]^2(0.1)$
  $[Ag^+]^2 = \frac{1.1 	imes 10^{-12}}{0.1} = 1.1 	imes 10^{-11}$
  $[Ag^+] = \sqrt{1.1 	imes 10^{-11}} \approx 3.3 	imes 10^{-6} M$
 3. The concentration of Ag$^+$ ions in the saturated solution is approximately $3.3 	imes 10^{-6}$ M.
  Result: $3.3 	imes 10^{-6}$ M
  Verification: Substitute the calculated $[Ag^+]$ back into the $K_{sp}$ expression to check if it matches the given value.
 ## OPTION ANALYSIS
A :x: Incorrect calculation of $[Ag^+]$.
B ✓: Correct calculation of $[Ag^+]$ using the common ion effect.
C :x: Incorrect calculation, likely a mistake in the square root operation or handling of the $K_{sp}$ value.
D :x: Incorrect calculation, possibly a factor of 2 error.
 ## LEARNING AIDS
 **Common Mistakes**:
  1.Ignoring the common ion effect: Students might calculate the solubility as if no chromate ions were already present.
  How to avoid: Remember that the presence of a common ion (CrO$_4^{2-}$ in this case) reduces the solubility of the sparingly soluble salt.
  2.Incorrect stoichiometry: Misinterpreting the stoichiometric coefficients in the equilibrium expression can lead to errors.
  How to avoid: Carefully write the balanced dissolution equation and derive the correct $K_{sp}$ expression.
 **Quick Tips**:
  1.Remember the common ion effect always decreases the solubility.
  2.Always write the balanced chemical equation before setting up the $K_{sp}$ expression.

Mr. S M HUSSAIN · Answer

1.1 x 10$^{-6}$ M

Mr. S M HUSSAIN · Answer

The correct Option is "3.3 x 10$^{-6}$ M"

Mr. S M HUSSAIN · Answer

1.1 x 10$^{-5}$ M

Mr. S M HUSSAIN · Answer

2.2 x 10$^{-5}$ M

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