Problem 92 - Introductory Chemistry Atoms First

Question

Consider a saturated aqueous solution of \(\mathrm{AgCl}\), a salt that is only sparingly soluble in water. What happens to this solution if a saturated solution of NaCl (a water-soluble salt) is added to it? (Hint: If \(\left[\mathrm{Ag}^{+}(a q)\right] \times\left[\mathrm{Cl}^{-}(a q)\right]>K_{\mathrm{sp}^{\prime}}\), precipitation will occur.)

Step-by-Step Solution

Answer

When a saturated solution of NaCl is added to a saturated aqueous solution of AgCl, the concentration of Cl⁻ ions increases, which increases the reaction quotient, Q. As a result, \(Q > K_{\mathrm{sp}^{\prime}}\), and precipitation occurs. Therefore, AgCl will start to come out of the solution as a solid.

Step 1: Write the dissolution equilibrium reactions for AgCl and NaCl

For AgCl, the equilibrium reaction is: \[ \mathrm{AgCl}(s) \rightleftharpoons \mathrm{Ag}^{+}(a q) + \mathrm{Cl}^{-}(a q) \] and for NaCl, the equilibrium reaction is: \[ \mathrm{NaCl}(s) \rightleftharpoons \mathrm{Na}^{+}(a q) + \mathrm{Cl}^{-}(a q) \]

Step 2: Determine the effect of adding NaCl to the AgCl solution

When we add NaCl to the saturated AgCl solution, the concentration of Cl- ions will increase. Since the reaction quotient Q of AgCl depends on the product of ion concentrations, we need to calculate the new values of Q and compare it to the Kₛₚ (\(K_{\mathrm{sp}^{\prime}}\)) of AgCl.

Step 3: Calculate the reaction quotient Q

The reaction quotient Q for the dissolution of AgCl can be expressed as: \[Q = [\mathrm{Ag}^{+}(a q)][\mathrm{Cl}^{-}(a q)]\] By adding NaCl to the solution, we increase the concentration of Cl- ions. As a result, the value of Q will increase.

Step 4: Compare Q to Kₛₚ

The solubility product constant, Kₛₚ, is a measure of the equilibrium between the dissolved ions and the undissolved salt. If the reaction quotient, Q, is: - Less than Kₛₚ, the reaction proceeds forward, and the salt will dissolve. - Equal to Kₛₚ, the reaction is at equilibrium, and no further dissolution or precipitation occurs. - Greater than Kₛₚ, the reaction proceeds in reverse, and precipitation occurs.

Step 5: Interpret the results in terms of precipitation or dissolution

Since the addition of NaCl increases the concentration of Cl- ions, and in turn, increases the value of Q, it is most likely that \(Q > K_{\mathrm{sp}^{\prime}}\). This means that precipitation will occur, and AgCl will start to come out of the solution as a solid.

Key Concepts

Precipitation Reaction

Imagine you’re adding a spoon of sugar to your tea, and instead of dissolving, it forms a clump at the bottom. That’s similar to what happens in a precipitation reaction but with chemicals. In a precipitation reaction, two soluble substances mix and form an insoluble substance, or a precipitate. For example, when a saturated solution of silver chloride (\r\(\mathrm{AgCl}\)) is mixed with a solution containing chloride ions (\r\(\mathrm{Cl}^{-}\)), the excessive chloride ions push the solution beyond its solubility limits, and solid silver chloride begins to form and settle out. This settling out process is known as precipitation.
\r
\rPrecipitation reactions are important for several reasons. They are fundamental in water treatment facilities to remove impurities, and they are also used in qualitative analysis to identify the presence of various ions in a solution. To predict whether a precipitation reaction will occur, one can refer to the solubility rules or to the concept of the solubility product.

Solubility Product Constant (Ksp)

The solubility product constant (Ksp) is your mathematical crystal ball to predict if a solid will stay dissolved or crash out as a precipitate. It’s a unique value for every sparingly soluble compound, like a signature. In our tea metaphor, think of it as the maximum amount of sugar your tea can hold before it says 'no more!'.
\r
\rFor the sparingly soluble salt, silver chloride (\r\(\mathrm{AgCl}\)), the equilibrium between solid \r\(\mathrm{AgCl}\) and its dissolved ions \r\(\mathrm{Ag}^{+}\) and \r\(\mathrm{Cl}^{-}\) can be written as:\r\[\mathrm{AgCl}(s) \rightleftharpoons \mathrm{Ag}^{+}(a q) + \mathrm{Cl}^{-}(a q)\]
\rThe \r\(K_{\text{sp}}\) is the product of the concentrations of these ions at equilibrium. If you know the \r\(K_{\text{sp}}\), you can figure out whether a solution is supersaturated and ready to precipitate or if it can dissolve more solid.

Common Ion Effect

Ever try talking in a crowded room? Your voice just drowns out. That’s the common ion effect, but with ions. It describes how the addition of an ion that’s already in the solution affects the solubility of the compound. In our exercise example, when we add a saturated solution of sodium chloride (\r\(\mathrm{NaCl}\)), which dissociates into \r\(\mathrm{Na}^{+}\) and \r\(\mathrm{Cl}^{-}\), we are increasing the number of \r\(\mathrm{Cl}^{-}\) ions already present from the dissociation of \r\(\mathrm{AgCl}\).
\r
\rThe solution can only handle so many \r\(\mathrm{Cl}^{-}\) ions before it's like our crowded room — and no more dissolution can occur. So, adding more of this common ion (\r\(\mathrm{Cl}^{-}\)) encourages the insoluble compound (\r\(\mathrm{AgCl}\)) to form, reducing the overall solubility of silver chloride in the solution. This effect helps to explain why we observe a precipitation reaction in the presence of a common ion.

Reaction Quotient (Q)

The reaction quotient (Q) is like a real-time scoreboard for a chemical reaction. It tells you what's currently happening in a solution. The value of Q is calculated the same way as \r\(K_{\text{sp}}\), but with the current, not equilibrium, concentrations of the ions involved.
\r
\rYou calculate Q by taking the product of the concentrations of the ions involved in the reaction at a given moment:\r\[Q = [\mathrm{Ag}^{+}(aq)][\mathrm{Cl}^{-}(aq)]\]
\rWhen you compare Q to the \r\(K_{\text{sp}}\) of the compound, you can predict what’s next for the reaction. If \r\(Q < K_{\text{sp}}\), the system will try to dissolve more solid to reach equilibrium. If \r\(Q = K_{\text{sp}}\), it is at equilibrium, and no net change occurs. But if \r\(Q > K_{\text{sp}}\), then hold onto your hats! — it means that there’s too much solute, and the excess is going to precipitate out. In the exercise, by adding NaCl, we raise the \r\(Q\) until it surpasses the \r\(K_{\text{sp}}\) for \r\(\mathrm{AgCl}\), leading to the conclusion that precipitation of \r\(\mathrm{AgCl}\) will occur.