Question

Strontium carbonate dissociates slightly in an aqueous solution as follows: $$ \mathrm{SrCO}_{3}(s) \rightleftarrows \mathrm{Sr}^{2+}(a q)+\mathrm{CO}_{3}^{2-}(a q) $$ Predict the direction of equilibrium shift for each of the following stresses: (a) increase \(\left[\mathrm{Sr}^{2+}\right]\) (b) increase \(\left[\mathrm{CO}_{3}{ }^{2-}\right]\) (c) decrease \(\left[\mathrm{Sr}^{2+}\right]\) (d) decrease \(\left[\mathrm{CO}_{3}^{2-}\right]\) (e) add solid \(\mathrm{SrCO}_{3}\) (f) add solid \(\operatorname{Sr}\left(\mathrm{NO}_{3}\right)_{2}\) (g) add solid \(\mathrm{KNO}_{3}\) (h) add \(\mathrm{H}^{+}\)

Short answer

The equilibrium shifts left with increased \([\mathrm{Sr}^{2+}]\) and \([\mathrm{CO}_{3}^{2-}]\) or added \(\mathrm{Sr}(\mathrm{NO}_{3})_{2}\); right with decreased concentrations or added \(\mathrm{H}^{+}\); unaffected by adding \(\mathrm{SrCO}_{3}\) or \(\mathrm{KNO}_{3}\).

Step-by-step solution

Understanding the Equilibrium Reaction

The dissociation of strontium carbonate in water is represented by the equilibrium: \(\mathrm{SrCO}_{3}(s) \rightleftarrows \mathrm{Sr}^{2+}(aq)+\mathrm{CO}_{3}^{2-}(aq)\). According to Le Chatelier's Principle, the system will shift in a direction that opposes any applied stress to restore equilibrium.

Analyzing Increase in \([\mathrm{Sr}^{2+}]\)

(a) An increase in the concentration of \([\mathrm{Sr}^{2+}]\) will cause the equilibrium to shift to the left, favoring the formation of \(\mathrm{SrCO}_{3}(s)\) to reduce the concentration of \([\mathrm{Sr}^{2+}]\).

Analyzing Increase in \([\mathrm{CO}_{3}^{2-}]\)

(b) Similarly, an increase in \([\mathrm{CO}_{3}^{2-}]\) will also shift the equilibrium to the left, forming more \(\mathrm{SrCO}_{3}(s)\) to decrease the \([\mathrm{CO}_{3}^{2-}]\) concentration.

Analyzing Decrease in \([\mathrm{Sr}^{2+}]\)

(c) A decrease in \([\mathrm{Sr}^{2+}]\) concentration will cause the equilibrium to shift to the right, dissociating more \(\mathrm{SrCO}_{3}(s)\) to increase \([\mathrm{Sr}^{2+}]\) concentration and restore balance.

Analyzing Decrease in \([\mathrm{CO}_{3}^{2-}]\)

(d) Similarly, a decrease in \([\mathrm{CO}_{3}^{2-}]\) will cause the equilibrium to shift to the right, producing more \([\mathrm{CO}_{3}^{2-}]\) from \(\mathrm{SrCO}_{3}(s)\).

Effect of Adding Solid \(\mathrm{SrCO}_{3}\)

(e) Adding more solid \(\mathrm{SrCO}_{3}\) will not change the position of the equilibrium because \(\mathrm{SrCO}_{3}\) is a solid, and changes in the concentration of solids do not affect the equilibrium.

Effect of Adding Solid \(\mathrm{Sr}(\mathrm{NO}_{3})_{2}\)

(f) Adding \(\mathrm{Sr}(\mathrm{NO}_{3})_{2}\) will increase \([\mathrm{Sr}^{2+}]\), causing the equilibrium to shift to the left to form more solid \(\mathrm{SrCO}_{3}\).

Effect of Adding Solid \(\mathrm{KNO}_{3}\)

(g) Adding \(\mathrm{KNO}_{3}\) will not affect the equilibrium position as neither \(\mathrm{K}^{+}\) nor \(\mathrm{NO}_{3}^{-}\) ions are involved in the equilibrium reaction.

Effect of Adding \(\mathrm{H}^{+}\)

(h) Adding \(\mathrm{H}^{+}\) ions will react with \(\mathrm{CO}_{3}^{2-}\) ions to form \(\mathrm{HCO}_{3}^{-}\) or \(\mathrm{H}_{2}\mathrm{CO}_{3}\). This decrease in \([\mathrm{CO}_{3}^{2-}]\) will cause the equilibrium to shift to the right to produce more \(\mathrm{CO}_{3}^{2-}\).

Key concepts

Equilibrium Reaction

In chemistry, an equilibrium reaction occurs when a reversible reaction has reached a point where the rate of the forward reaction equals the rate of the reverse reaction. This means the concentrations of reactants and products remain constant over time, although not necessarily equal. The concept of equilibrium is crucial, as it helps to understand how conditions such as concentration, pressure, and temperature can affect chemical reactions. For strontium carbonate, the equilibrium reaction can be represented as: \[ \mathrm{SrCO}_{3}(s) \rightleftharpoons \mathrm{Sr}^{2+}(aq) + \mathrm{CO}_{3}^{2-}(aq) \]This reaction shows how solid strontium carbonate can dissolve into its aqueous ions, forming an equilibrium between the undissolved solid and its dissolved ions in the solution.

Le Chatelier's Principle is an important guideline that helps predict how the equilibrium will respond to disturbances or "stresses." The principle dictates that when a system at equilibrium is subjected to a change in concentration, pressure, or temperature, the system will adjust by shifting the equilibrium to counteract the effect of the disturbance.

Strontium Carbonate

Strontium carbonate, represented as \(\mathrm{SrCO}_{3}\), is an ionic compound composed of strontium ions \(\mathrm{Sr}^{2+}\) and carbonate ions \(\mathrm{CO}_{3}^{2-}\). This white crystalline solid is fairly insoluble in water, which means it doesn't dissolve easily, but in an equilibrium reaction, a small amount can dissociate into its ions.

Understanding this compound is important for scenarios like mineral extraction, ceramics, and some industrial applications. When studying its behavior in an equilibrium reaction, the focus is on the dynamic balance between the solid form and its ionic form in solution. Since it is slightly soluble, changes such as adding or removing ions can influence which side of the equilibrium is favored, illustrating Le Chatelier's Principle in action.

• This is why increasing the concentration of \( \mathrm{Sr}^{2+} \) or \( \mathrm{CO}_{3}^{2-} \) shifts the equilibrium towards forming more solid \( \mathrm{SrCO}_{3} \).
• Conversely, a decrease in these ions will tend to promote more dissolution to restore the equilibrium balance.

Aqueous Solution

An aqueous solution is one where the solvent is water. In this context, it means that when strontium carbonate dissociates, it does so in water, producing aqueous ions like \(\mathrm{Sr}^{2+}(aq)\) and \(\mathrm{CO}_{3}^{2-}(aq)\). These ions are free to move in the solution, which makes water a versatile medium facilitating various chemical reactions. Aqueous solutions play a key role in chemical equilibrium because the concentration of ions in solution is pivotal in determining the position of the equilibrium.

• In the case of \(\mathrm{SrCO}_{3} \), when additional amounts of acidic protons \(\mathrm{H}^{+}\) are introduced, they typically interact with \(\mathrm{CO}_{3}^{2-} \), affecting the ionic balance.
• This interaction demonstrates how changes in solution composition can drive the direction in which the equilibrium will shift, illustrating the 'opposing change' characteristic of Le Chatelier's Principle.
• Other ions that don't directly participate in the equilibrium, such as \(\mathrm{K}^{+}\) or \(\mathrm{NO}_{3}^{-}\), may be present but do not perturb the established equilibrium because they don't affect the concentrations of the reacting species.