Given the equilibrium:
HCN(aq) + H2O(l) ←→ H3O+(aq) + CN-(aq) ΔH >0; Ka = 4.0 x10-10
What happens to the concentration of hydrogen cyanide [HCN] when the following stresses are
placed on the system at equilibrium?
NaCl is added ---> Stays the Same but Why??
NaOH is added---> decreases, but why??
NaCN is added---> increases, but why doesn't it decrease? I thought it might add to CN^- ??
HCl is added---> increases, why does this make HCN increase while NaCl doesn't affect it??
I understood the other given stresses so I didn't list them, but I am pretty confused about why these answers are correct and not others...any help breaking this down is appreciated--thank you!!
Val, I have never understood exactly why Le Chatlier's Prinicple becomes difficult for students. The principle is simply one of shifting the equilibrium to UNDO what has been done.
HCN(aq) + H2O(l) ←→ H3O+(aq) + CN-(aq) ΔH >0; Ka = 4.0 x10-10
What happens to the concentration of hydrogen cyanide [HCN] when the following stresses are
placed on the system at equilibrium?
NaCl is added ---> Stays the Same but Why??
NaCl. Neither the Na^+ nor the Cl^- hydrolyzes(reacts with water); therefore, neither H3O^+ nor CN^- are affected. No change
NaOH is added---> decreases, but why??
NaOH is a strong base, it neutralizes H3O^+ on the right side of the equation, that removes H3O^+ from the right side, the reaction REPLACES H3O^+ on the right side to make up for (UNDOES) what has been lost, the only way it can do that is for HCN to react with H2O to produce H3O^+ and that DECREASES HCN. Right?
NaCN is added---> increases, but why doesn't it decrease? I thought it might add to CN^- ??
Increasing CN^- on the right, the reaction must undo what has been done so it must try to decrease CN^-. How can it do that. The CN^- must react with H3O^+ to produce HCN (the reaction must shift to the left) so HCN increases.
HCl is added---> increases, why does this make HCN increase while NaCl doesn't affect it??
With NaCl, neither Na^+ nor Cl^- is involved in the equilibrium so the HCN + H2O reaction could care less about how much NaCl is dumped in. But H3O^+ IS involved and the reaction does care. Adding HCl means we are adding H3O^+, the reaction must undo that addition, the only way it can do that is to react H3O^+ with CN^- to produce HCN and H2O (the reaction must shift to the left). That means more HCN is produced.
You need not go through all of that long list of reasoning each time. Here is a simple way to know which way the reaction shifts. (And truthfully, if you know which way it shifts you know which reagents are increased and which are decreased).
The reaction shifts AWAY FROM what you did. Try it? CN^- added. Moves to the left with products decreasing reactants increasing. H3O^+ added. Moves to the left with products decreasing and reactants increasing. H2O added. Moves to the right with reactants decreasing and products increasing. That's really all you need to remember. The reaction moves AWAY FROM what you did. One caveat here is not to get caught up in all of the numbers. For example, adding NaCN. That addes CN^-. The reaction shifts to the left. HCN is increased. H2O is increased. H3O^+ is decreased. So far, so good. CN^- is decreased, too, FROM ITS ORIGINAL CONCENTRATION (that is, some of what was there at the beginning moves to the left) but we ADDED CN^- so the overall concn CN^- MUST increase. To put numbers on it, if we started with 0.1 M CN^- in the reaction at equilibrium and we added 0.1 M CN^- to it, the 0.1 that was there to begin with moves to say 0.08 or 0.05 but the 0.1 we ADDED makes the total something like 0.15-0.18 (but not 0.2). I hope this takes care of your problem. If not, post away.
Ah, equilibrium and stresses, always a fun topic! Let's dive in and see if we can make some sense of it.
When NaCl is added, the concentration of hydrogen cyanide [HCN] stays the same. Why? Well, adding NaCl doesn't directly affect the concentration of HCN or any other species in the equilibrium expression. It's basically like adding a spectator at a magic show – it doesn't really change what's happening on stage.
Now, when NaOH is added, the concentration of [HCN] actually decreases. This can be explained by Le Chatelier's principle. Adding NaOH introduces hydroxide ions (OH-) into the system, which react with H3O+ to produce more water. This shifts the equilibrium to the left, decreasing the concentration of HCN.
Things get a bit spicier when NaCN is added. You're right in thinking that NaCN would add to the concentration of CN-. However, this doesn't lead to a decrease in [HCN] because HCN and CN- are not in a direct 1:1 relationship. HCN is the acid, while CN- is the conjugate base. By adding CN-, the equilibrium will shift to the right to reestablish the balance, resulting in an increase in [HCN].
Finally, when HCl is added, the concentration of HCN increases. HCl is a strong acid, so it donates a proton to the reaction. This increase in H3O+ concentration pushes the equilibrium to the right in order to reduce the excess protons, leading to an increase in [HCN].
I hope this breakdown helps make things clearer for you! Remember, chemistry can be a bit of a clown sometimes – it's all about finding the right balance.
Let's break down each of the given stresses on the equilibrium system and understand their effects on the concentration of hydrogen cyanide ([HCN]).
1. NaCl is added: This stress does not affect the concentration of hydrogen cyanide ([HCN]). Adding NaCl does not introduce any new species, ions, or reactants that directly interact with or affect the HCN equilibrium. Therefore, the concentration of HCN remains the same.
2. NaOH is added: The addition of NaOH causes the hydroxide ions (OH-) to react with the hydronium ions (H3O+) in the equilibrium. This reaction consumes H3O+, shifting the equilibrium towards the left to restore the balance. As a result, the concentration of HCN decreases.
3. NaCN is added: The addition of NaCN introduces cyanide ions (CN-) into the equilibrium. In the equilibrium, CN- is already a product and increasing its concentration will shift the equilibrium towards the reactants to counterbalance the change. However, this will only have a minor effect on the concentration of HCN because the value of the equilibrium constant (Ka) is very small (4.0 x 10^-10). Therefore, the concentration of HCN will increase by a small amount.
4. HCl is added: Adding HCl introduces chloride ions (Cl-) into the equilibrium system. The chloride ions do not directly interact with the HCN equilibrium. However, the additional HCl increases the concentration of the hydronium ions (H3O+), the reactant side of the equilibrium. This increase in H3O+ concentration drives the equilibrium towards the products, causing the concentration of HCN to increase.
In summary:
- Adding NaCl has no direct effect on the concentration of HCN.
- Adding NaOH decreases the concentration of HCN.
- Adding NaCN increases HCN concentration (although only by a small amount due to the small value of Ka).
- Adding HCl increases the concentration of HCN.
To determine what happens to the concentration of hydrogen cyanide [HCN] when different stresses are placed on the system at equilibrium, we need to consider Le Chatelier's principle. Le Chatelier's principle states that if a system at equilibrium is subjected to a stress, the system will adjust to minimize the effect of that stress and restore equilibrium.
1. When NaCl is added, the concentration of hydrogen cyanide [HCN] stays the same. This is because NaCl is a neutral compound that does not affect the concentration of HCN or the equilibrium. The addition of NaCl does not introduce any ions that could react with or shift the equilibrium of the HCN dissociation reaction. Therefore, the concentration of HCN is not affected.
2. When NaOH is added, the concentration of hydrogen cyanide [HCN] decreases. This is because NaOH is a strong base that reacts with HCN to form additional H2O and the corresponding cyanide ion (CN-). The reaction can be written as:
HCN(aq) + NaOH(aq) → H2O(l) + NaCN(aq)
Since the reaction consumes HCN and produces more products (H2O and NaCN), the equilibrium shifts to the left to minimize the effect of the added base. As a result, the concentration of HCN decreases.
3. When NaCN is added, the concentration of hydrogen cyanide [HCN] increases, and it does not decrease. This may seem counterintuitive since NaCN contains the cyanide ion (CN-), which is a product of the dissociation reaction. However, Le Chatelier's principle tells us that if a reactant is added to the system, the equilibrium will shift to minimize the effect of that added reactant.
In this case, the addition of NaCN introduces additional CN- ions, which are a product of the reaction. To restore equilibrium, the reaction will shift to the right, resulting in the formation of more HCN to offset the increase in CN- concentration. Consequently, the concentration of HCN increases.
4. When HCl is added, the concentration of hydrogen cyanide [HCN] increases. HCl is a strong acid that reacts with water to produce additional H3O+ ions. The reaction can be written as:
HCl(aq) + H2O(l) → H3O+(aq) + Cl-(aq)
Since the reaction produces H3O+ ions, the equilibrium of the HCN dissociation reaction is affected, and it shifts to the right to minimize the effect of the added acid. As a result, more HCN is dissociated, leading to an increase in its concentration.
In summary, the addition of NaCl does not affect HCN concentration because it is a neutral compound. The addition of NaOH causes a decrease in HCN concentration due to the formation of additional products. The addition of NaCN results in an increase in HCN concentration as equilibrium shifts to consume the added product. Finally, the addition of HCl causes an increase in HCN concentration due to the consumption of H3O+ ions.