To determine the effect on the equilibrium when the pH is increased, we must consider how the system will respond according to Le Chatelier's principle. Le Chatelier’s principle states that a system in equilibrium will adjust to counteract any changes imposed on it.
We are given the following chemical equilibrium reactions:
[tex]\[ CO _{2(g)}+ H _2 O _{(l)} \longleftrightarrow H _2 CO _{3(aq)} \][/tex]
[tex]\[ H _2 CO _{3(aq)} \longleftrightarrow H ^{+}(aq)+ HCO _3^{-}(aq) \][/tex]
Step-by-Step Solution:
1. Understanding pH Increase:
- Increasing the pH of a solution means reducing the concentration of hydrogen ions ([tex]\( H^+ \)[/tex]) in the solution, making it less acidic.
2. Response to the Change in [tex]\( H^+ \)[/tex] Concentration:
- According to Le Chatelier's principle, the equilibrium will shift to counteract the decrease in [tex]\( H^+ \)[/tex] ions. To increase the [tex]\( H^+ \)[/tex] ion concentration, the system will favor the direction that produces more [tex]\( H^+ \)[/tex].
3. Identifying Affected Equilibria:
- The second equation directly involves [tex]\( H_2CO_3 \)[/tex] dissociating into [tex]\( H^+ \)[/tex] and [tex]\( HCO_3^- \)[/tex]:
[tex]\[ H_2CO_3(aq) \longleftrightarrow H^+(aq) + HCO_3^-(aq) \][/tex]
- To counteract the reduction in [tex]\( H^+ \)[/tex], the reverse reaction (formation of [tex]\( H_2CO_3 \)[/tex] from [tex]\( H^+ \)[/tex] and [tex]\( HCO_3^- \)[/tex]) will be favored.
4. Effect on the First Equation:
- As more [tex]\( H_2CO_3 \)[/tex] is formed, the equilibrium of the first reaction will also be affected:
[tex]\[ CO_2(g) + H_2O(l) \longleftrightarrow H_2CO_3(aq) \][/tex]
- With more [tex]\( H_2CO_3 \)[/tex] being in demand to balance the [tex]\( H^+ \)[/tex] ions being consumed, this equilibrium will shift towards producing more [tex]\( H_2CO_3 \)[/tex] from [tex]\( CO_2 \)[/tex] and [tex]\( H_2O \)[/tex].
5. Final Conclusion:
- The overall effect of increasing the pH (reducing [tex]\( H^+ \)[/tex] concentration) will lead the system to produce more [tex]\( H_2CO_3 \)[/tex] to counteract the loss of [tex]\( H^+ \)[/tex] ions.
Therefore, the equilibrium will shift such that:
More [tex]\( H_2CO_3 \)[/tex] will be produced.
This detailed analysis leads us to conclude that the correct answer is:
- More [tex]\( H_2CO_3 (aq) \)[/tex] will be produced.
