Answer :
To determine which compound will cause a shift in the equilibrium of the reaction:
[tex]\[ CaCO_3(s) \Leftrightarrow Ca^{2+}(aq) + CO_3^{2-}(aq) \][/tex]
we need to consider how each compound might affect the concentrations of the products or reactants and subsequently how the equilibrium will respond according to Le Chatelier's principle.
1. [tex]\(CCl_4\)[/tex]: This compound is carbon tetrachloride and it is nonpolar and does not dissolve in water. It has no bearing on the ionic species in the solution, so it will not affect the equilibrium.
2. [tex]\(CO_2\)[/tex]: Carbon dioxide, when dissolved in water, forms carbonic acid ([tex]\(H_2CO_3\)[/tex]). The dissociation of carbonic acid can lead to the formation of bicarbonate ([tex]\(HCO_3^-\)[/tex]) and subsequently carbonate ions ([tex]\(CO_3^{2-}\)[/tex]). However, the direct addition of [tex]\(CO_2\)[/tex] does not immediately affect the concentration of [tex]\(CO_3^{2-}\)[/tex] ions required for a significant impact on our system.
3. [tex]\(CuSO_4\)[/tex]: Copper(II) sulfate will dissociate into [tex]\(Cu^{2+}\)[/tex] and [tex]\(SO_4^{2-}\)[/tex] ions in solution. Neither of these ions directly affects the concentration of [tex]\(Ca^{2+}\)[/tex] or [tex]\(CO_3^{2-}\)[/tex] in a way that relates to our equilibrium system.
4. [tex]\(Na_2CO_3\)[/tex]: Sodium carbonate dissociates completely in water to form [tex]\(2 Na^+\)[/tex] and [tex]\(CO_3^{2-}\)[/tex] ions. Adding [tex]\(Na_2CO_3\)[/tex] will increase the concentration of [tex]\(CO_3^{2-}\)[/tex] in the solution. According to Le Chatelier's principle, if the concentration of [tex]\(CO_3^{2-}\)[/tex] increases, the equilibrium will shift to the left, to produce more [tex]\(CaCO_3(s)\)[/tex], and thus decrease the concentration of [tex]\(CO_3^{2-}\)[/tex].
Therefore, the addition of [tex]\(Na_2CO_3\)[/tex] will cause a shift in the equilibrium of the reaction:
[tex]\[ \boxed{Na_2CO_3} \][/tex]
is the compound that, when added, will cause a shift in equilibrium.
[tex]\[ CaCO_3(s) \Leftrightarrow Ca^{2+}(aq) + CO_3^{2-}(aq) \][/tex]
we need to consider how each compound might affect the concentrations of the products or reactants and subsequently how the equilibrium will respond according to Le Chatelier's principle.
1. [tex]\(CCl_4\)[/tex]: This compound is carbon tetrachloride and it is nonpolar and does not dissolve in water. It has no bearing on the ionic species in the solution, so it will not affect the equilibrium.
2. [tex]\(CO_2\)[/tex]: Carbon dioxide, when dissolved in water, forms carbonic acid ([tex]\(H_2CO_3\)[/tex]). The dissociation of carbonic acid can lead to the formation of bicarbonate ([tex]\(HCO_3^-\)[/tex]) and subsequently carbonate ions ([tex]\(CO_3^{2-}\)[/tex]). However, the direct addition of [tex]\(CO_2\)[/tex] does not immediately affect the concentration of [tex]\(CO_3^{2-}\)[/tex] ions required for a significant impact on our system.
3. [tex]\(CuSO_4\)[/tex]: Copper(II) sulfate will dissociate into [tex]\(Cu^{2+}\)[/tex] and [tex]\(SO_4^{2-}\)[/tex] ions in solution. Neither of these ions directly affects the concentration of [tex]\(Ca^{2+}\)[/tex] or [tex]\(CO_3^{2-}\)[/tex] in a way that relates to our equilibrium system.
4. [tex]\(Na_2CO_3\)[/tex]: Sodium carbonate dissociates completely in water to form [tex]\(2 Na^+\)[/tex] and [tex]\(CO_3^{2-}\)[/tex] ions. Adding [tex]\(Na_2CO_3\)[/tex] will increase the concentration of [tex]\(CO_3^{2-}\)[/tex] in the solution. According to Le Chatelier's principle, if the concentration of [tex]\(CO_3^{2-}\)[/tex] increases, the equilibrium will shift to the left, to produce more [tex]\(CaCO_3(s)\)[/tex], and thus decrease the concentration of [tex]\(CO_3^{2-}\)[/tex].
Therefore, the addition of [tex]\(Na_2CO_3\)[/tex] will cause a shift in the equilibrium of the reaction:
[tex]\[ \boxed{Na_2CO_3} \][/tex]
is the compound that, when added, will cause a shift in equilibrium.
