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March 24, 2017

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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!!

  • Chemistry--help! - ,

    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.

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