Problem 113 - Chemistry: An Atoms-Focused Approach

Question

When silver oxide dissolves in water, the following reaction occurs: $$ \mathrm{Ag}_{2} \mathrm{O}(s)+\mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow 2 \mathrm{Ag}^{+}(a q)+2 \mathrm{OH}^{-}(a q) $$ If a saturated aqueous solution of silver oxide is \(1.6 \times 10^{-4} M\) in hydroxide ion, what is the \(K_{\mathrm{sp}}\) of silver oxide?

Step-by-Step Solution

Answer

Answer: The Ksp of silver oxide (Ag2O) is approximately \(6.55 \times 10^{-16}\).

Step 1: Identify the relationship between the Ag+ and OH- concentrations in the solution

Since 1 mol of Ag2O produces 2 moles of Ag+ and 2 moles of OH- following the balanced equation, the concentration of Ag+ is equal to 2 times the concentration of dissolved silver oxide. Let x be the molar concentration of dissolved Ag2O. Then the concentration of Ag+ is 2x and the concentration of OH- is 2x.

Step 2: Write the equilibrium expression for the Ksp of Ag2O

Using the balanced equation: $$ \mathrm{Ag}_{2} \mathrm{O}(s)+\mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow 2 \mathrm{Ag}^{+}(a q)+2 \mathrm{OH}^{-}(a q) $$ The Ksp of Ag2O can be written as follows: $$ K_{sp} = [\mathrm{Ag}^{+}]^2 [\mathrm{OH}^{-}]^2 $$

Step 3: Substitute the given OH- concentration to solve for the Ksp

The concentration of OH- in the saturated solution is given as \(1.6 \times 10^{-4} \text{M}\). This means that 2x = \(1.6 \times 10^{-4}\), so x = \(0.8 \times 10^{-4}\). The concentration of Ag+ is also 2x, which is equal to \(1.6 \times 10^{-4} \text{M}\). We can substitute these values into the Ksp expression: $$ K_{sp} = (1.6 \times 10^{-4})^2 (1.6 \times 10^{-4})^2 $$

Step 4: Calculate the Ksp for Ag2O

Now we can calculate the Ksp value: $$ K_{sp} = (1.6 \times 10^{-4})^2 (1.6 \times 10^{-4})^2 = 6.5536 \times 10^{-16} $$ The Ksp of silver oxide (Ag2O) is approximately \(6.55 \times 10^{-16}\).