Question

Concentration of the $Ag^+$ ions in a saturated solution of $Ag_2C_2O_4$ is $2.2 \times 10^{-4}\, mol\, L^{-1}$ Solubility product of $Ag_2C_2O_4$ is :-

• A. $2.66 \times 10^{-12}$
• B. $4.5 \times 10^{-11}$
• C. $5.3 \times 10^{-12}$
• D. $2.42 \times 10^{-8}$

Answer 1

The Correct Option is C

 To solve this problem, we need to calculate the solubility product (K_sp) of silver oxalate, Ag₂C₂O₄, from the given concentration of Ag⁺ ions.

The dissolution of Ag₂C₂O₄ in water can be represented by the following equilibrium equation:

Ag_2C_2O_4(s) \rightleftharpoons 2Ag^+(aq) + C_2O_4^{2-}(aq)

Let the solubility of Ag₂C₂O₄ be \(s\) mol/L. For each mole of Ag₂C₂O₄ that dissolves, 2 moles of Ag⁺ and 1 mole of C₂O₄^2- are produced. Therefore:

• Concentration of Ag⁺ ions = \(2s\)
• Concentration of C₂O₄^2- ions = \(s\)

We are given the concentration of Ag⁺ ions as \(2.2 \times 10^{-4}\, mol\, L^{-1}\). Therefore, we can write:

2s = 2.2 \times 10^{-4}

Solving for \(s\) gives:

s = \frac{2.2 \times 10^{-4}}{2} = 1.1 \times 10^{-4}\, mol\, L^{-1}

Now, we can calculate the solubility product K_sp:

K_{sp} = [Ag^+]^2 \times [C_2O_4^{2-}]

K_{sp} = (2s)^2 \times s

K_{sp} = (2.2 \times 10^{-4})^2 \times 1.1 \times 10^{-4}

Calculate K_sp:

K_{sp} = 4.84 \times 10^{-8} \times 1.1 \times 10^{-4}

K_{sp} = 5.324 \times 10^{-12}

Rounding to two significant figures gives us the solubility product as:

K_{sp} = 5.3 \times 10^{-12}

Thus, the correct answer is \(5.3 \times 10^{-12}\).