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There are several chemical equilibria taking place in a solution of carbon dioxide in water: 1. CO₂(g) = CO₂(aq) 2. CO₂(aq) + H₂O(l) = H₂CO3(aq) 3. H₂CO3(aq) HCO3(aq) + H*(aq) 4. HCO3(aq) CO3² (aq) + H+ (aq) If OH- (from a strong base such as sodium hydroxide) is added to a solution of carbon dioxide, what happens to the OH and what effect will this have on the equilibria above? Select one: O O OH-(aq) reacts with the H*(aq) ions to form water (which is only very slightly dissociated). This shifts the third and fourth reactions to the right. However, there is no H*(aq) involved in either the first or second reactions, so these equilibria will remain unaffected by the additon of OH(aq). OH-(aq) is not involved in any of the four equilibrium reactions that carbon dioxide participates in and so the OH(aq) will have no influence on the carbon dioxide equilibria Carbon dioxide dissolves and then reacts to produce the acidic carbonic acid molecule (H₂CO3). The H₂CO3 molecule is only present in acidic solutions and not in basic solutions. This means that added OH-(aq) which makes the solution basic, will stop the H2CO3 molecule from forming and so the first and second equilibrium reactions above will be pushed to the left and less carbon dioxide will dissolve in the solution. OH-(aq) reacts with the H*(aq) ions to form water (which is only very slightly dissociated). This shifts the third and fourth reactions to the right which reduces the concentration of carbonic acid, which shifts the second reaction to the right, and then the first reaction shifts to the right so more CO₂(g) dissolves.

There are several chemical equilibria taking place in a solution of carbon dioxide in water: 1. CO₂(g) = CO₂(aq) 2. CO₂(aq) + H₂O(l) = H₂CO3(aq) 3. H₂CO3(aq) HCO3(aq) + H*(aq) 4. HCO3(aq) CO3² (aq) + H+ (aq) If OH- (from a strong base such as sodium hydroxide) is added to a solution of carbon dioxide, what happens to the OH and what effect will this have on the equilibria above? Select one: O O OH-(aq) reacts with the H*(aq) ions to form water (which is only very slightly dissociated). This shifts the third and fourth reactions to the right. However, there is no H*(aq) involved in either the first or second reactions, so these equilibria will remain unaffected by the additon of OH(aq). OH-(aq) is not involved in any of the four equilibrium reactions that carbon dioxide participates in and so the OH(aq) will have no influence on the carbon dioxide equilibria Carbon dioxide dissolves and then reacts to produce the acidic carbonic acid molecule (H₂CO3). The H₂CO3 molecule is only present in acidic solutions and not in basic solutions. This means that added OH-(aq) which makes the solution basic, will stop the H2CO3 molecule from forming and so the first and second equilibrium reactions above will be pushed to the left and less carbon dioxide will dissolve in the solution. OH-(aq) reacts with the H*(aq) ions to form water (which is only very slightly dissociated). This shifts the third and fourth reactions to the right which reduces the concentration of carbonic acid, which shifts the second reaction to the right, and then the first reaction shifts to the right so more CO₂(g) dissolves.

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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There are several chemical equilibria taking place in a solution of carbon dioxide in water: 1. CO₂(g) + CO₂(aq) 2. CO₂(aq) + H₂O(l) = H₂CO3(aq) 3. H₂CO3(aq) HCO3(aq) + H+ (aq) 4. HCO3(aq) = CO3²-(aq) + H*(aq) If OH- (from a strong base such as sodium hydroxide) is added to a solution of carbon dioxide, what happens to the OH- and what effect will this have on the equilibria above? Select one: OH-(aq) reacts with the H*(aq) ions to form water (which is only very slightly dissociated). This shifts the third and fourth reactions to the right. However, there is no H*(aq) involved in either the first or second reactions, so these equilibria will remain unaffected by the additon of OH¯(aq). O OH-(aq) is not involved in any of the four equilibrium reactions that carbon dioxide participates in and so the OH-(aq) will have no influence on the carbon dioxide equilibria Carbon dioxide dissolves and then reacts to produce the acidic carbonic acid molecule (H₂CO3). The H₂CO3 molecule is only present in acidic solutions and not in basic solutions. This means that added OH(aq) which makes the solution basic, will stop the H₂CO3 molecule from forming and so the first and second equilibrium reactions above will be pushed to the left and less carbon dioxide will dissolve in the solution. OH-(aq) reacts with the H*(aq) ions to form water (which is only very slightly dissociated). This shifts the third and fourth reactions to the right which reduces the concentration of carbonic acid, which shifts the second reaction to the right, and then the first reaction shifts to the right so more CO₂(g) dissolves.
Transcribed Image Text:There are several chemical equilibria taking place in a solution of carbon dioxide in water: 1. CO₂(g) + CO₂(aq) 2. CO₂(aq) + H₂O(l) = H₂CO3(aq) 3. H₂CO3(aq) HCO3(aq) + H+ (aq) 4. HCO3(aq) = CO3²-(aq) + H*(aq) If OH- (from a strong base such as sodium hydroxide) is added to a solution of carbon dioxide, what happens to the OH- and what effect will this have on the equilibria above? Select one: OH-(aq) reacts with the H*(aq) ions to form water (which is only very slightly dissociated). This shifts the third and fourth reactions to the right. However, there is no H*(aq) involved in either the first or second reactions, so these equilibria will remain unaffected by the additon of OH¯(aq). O OH-(aq) is not involved in any of the four equilibrium reactions that carbon dioxide participates in and so the OH-(aq) will have no influence on the carbon dioxide equilibria Carbon dioxide dissolves and then reacts to produce the acidic carbonic acid molecule (H₂CO3). The H₂CO3 molecule is only present in acidic solutions and not in basic solutions. This means that added OH(aq) which makes the solution basic, will stop the H₂CO3 molecule from forming and so the first and second equilibrium reactions above will be pushed to the left and less carbon dioxide will dissolve in the solution. OH-(aq) reacts with the H*(aq) ions to form water (which is only very slightly dissociated). This shifts the third and fourth reactions to the right which reduces the concentration of carbonic acid, which shifts the second reaction to the right, and then the first reaction shifts to the right so more CO₂(g) dissolves.
In part B of this experiment carbon dioxide gas was bubbled through a lime water (saturated calcium hydroxide) solution for several minutes. At Step 8 the lime water solution was heated on a steam bath for about three minutes and a precipitate was observed to form. From the options below, which give the equation(s) that accurate explain the observed results? Select one: O a. Ca²+ (aq) + CO3²-(aq) CaCO3(s) O b. COz(aq) → CO2(g) then CO₂(aq) + Ca²+ (aq) + 2HO (aq) = CaCO3(s) + H₂O(aq) O c. CO₂(aq) → CO₂ (s) O d. CO₂(aq) → CO₂(g) then CaCO3(aq) + H₂O(aq) Ca(OH)2(s) + CO₂(aq) Oe. CO₂(aq) = CO₂(g) then Ca²+ (aq) + 2HCO3(aq) = CaCO3(s) + CO₂(aq) + H₂O(aq)
Transcribed Image Text:In part B of this experiment carbon dioxide gas was bubbled through a lime water (saturated calcium hydroxide) solution for several minutes. At Step 8 the lime water solution was heated on a steam bath for about three minutes and a precipitate was observed to form. From the options below, which give the equation(s) that accurate explain the observed results? Select one: O a. Ca²+ (aq) + CO3²-(aq) CaCO3(s) O b. COz(aq) → CO2(g) then CO₂(aq) + Ca²+ (aq) + 2HO (aq) = CaCO3(s) + H₂O(aq) O c. CO₂(aq) → CO₂ (s) O d. CO₂(aq) → CO₂(g) then CaCO3(aq) + H₂O(aq) Ca(OH)2(s) + CO₂(aq) Oe. CO₂(aq) = CO₂(g) then Ca²+ (aq) + 2HCO3(aq) = CaCO3(s) + CO₂(aq) + H₂O(aq)
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