A saturated solution of silver trioxocarbonate (IV), was found to have concentration of 1.30 ×

Explanation

This question is testing your knowledge of solubility products. Solubility product (K\(_{sp}\)) is the equilibrium constant for a solid substance dissolving in an aqueous solution. It tells us about the maximum amount of a substance that can dissolve in a solution at equilibrium. The higher the K\(_{sp}\) value, the more soluble the substance is. In this question, we are given the concentration of a saturated solution of silver trioxocarbonate (IV), which is Ag\(_2\)CO\(_3\). The concentration is 1.30 \(\times\) 10\(^{-5}\) moldm\(^{-3}\). This means that the maximum amount of Ag\(_2\)CO\(_3\) that can dissolve in the solution has already dissolved and we have reached equilibrium. We are asked to find the solubility product (K\(_{sp}\)) of Ag\(_2\)CO\(_3\). To do this, we need to know the balanced chemical equation for the dissolution of Ag\(_2\)CO\(_3\): Ag\(_2\)CO\(_3\) \(\rightleftharpoons\) 2Ag\(^+\) + CO\(_3^{2-}\) The K\(_{sp}\) expression for Ag\(_2\)CO\(_3\) is then given by: K\(_{sp}\) = [Ag\(^+\)]\(^2\)[CO\(_3^{2-}\)] Since the concentration of Ag\(^+\) and CO\(_3^{2-}\) in the solution are equal (2:1 stoichiometric ratio), we can write: [Ag\(^+\)] = 2x and [CO\(_3^{2-}\)] = x Substituting these values into the K\(_{sp}\) expression, we get: K\(_{sp}\) = (2x)\(^2\)(x) = 4x\(^3\) Now, we can solve for x using the given concentration of the saturated solution: 1.30 \(\times\) 10\(^{-5}\) = 2x (since [Ag\(^+\)] = 2x) x = 6.50 \(\times\) 10\(^{-6}\) Substituting this value of x into the K\(_{sp}\) expression, we get: K\(_{sp}\) = 4x\(^3\) = 4(6.50 \(\times\) 10\(^{-6}\))\(^3\) = 8.79 \(\times\) 10\(^{-15}\) Therefore, the correct answer is Option A: 8.79 \(\times\) 10\(^{-15}\).