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Why do we need three equations to find the pH of NaCN, given Ka(HCN)?
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Lisa Aut
Why do we need three equations to find the pH of NaCN, given Ka(HCN)?
First and Third (sic!) equations are necessary. The second one just says "when I say that 1M solution of NaCN was formed, understand it as 1M of Na+ and 1M of CN- were added to a system".
Your key equilibrium is [HA]=[H+][A-] and [HA]/[H+][A-]=Ka. You then say HA+A- = total_molarity, [HA]=X, [H+]=X, [A-] = total-molarity - X and
x/([x]*[ total-molarity - X] = Ka. You know Ka, total_molarity. You now need to find x. In case of NaCN HCN is HA, CN- is A-.
Just plug your concentrations in the formula and get the result. Quite boring, indeed.
First and Third (sic!) equations are necessary. The second one just says "when I say that 1M solution of NaCN was formed, understand it as 1M of Na+ and 1M of CN- were added to a system".
Your key equilibrium is [HA]=[H+][A-] and [HA]/[H+][A-]=Ka. You then say HA+A- = total_molarity, [HA]=X, [H+]=X, [A-] = total-molarity - X and
x/([x]*[ total-molarity - X] = Ka. You know Ka, total_molarity. You now need to find x. In case of NaCN HCN is HA, CN- is A-.
Just plug your concentrations in the formula and get the result. Quite boring, indeed.
So strictly speaking you only really need equation 3 to solve this problem, but the first two equations help you figure out how to solve it if you're not too familiar. What do I mean:
Equation 1: $\ce{HCN + H2O <=> H3O+ + CN-}$
As you correctly note, this is to remind you of the definition for $K_a$
Equation 2: $\ce{NaCN -> Na+ + CN-}$
Note that I've changed $\ce{<=>}$ to $\ce{->}$. This is because most sodium salts fully dissociate in aqueous solution and so this "equilibrium" doesn't exist and you can then consider all $\ce{NaCN}$ to be converted to $\ce{CN-}$ for your "initial" stage in the ICE table. Hence there's no need to write an ICE Table with respect to Equation 2.
Value of writing this out:
In the case that you were given something sparingly soluble or did not fully dissociate, you would have to consider another equilibrium and this equation 2 would become significant. But in this case, $\ce{Na+}$ is a spectator ion which you can ignore.
Where does Equation 3 come from?
Equation 3 comes from the question itself (or rather what is happening in the system described). Remember, you've got a 1.0M solution of $\ce{NaCN}$. Since we've already established that it fully dissociates, at the "initial" stage of the system you have only $\ce{Na+}$ and $\ce{CN-}$ ions floating around. But from Equation 1 you know that $\ce{HCN}$ is a weak acid and so free $\ce{CN-}$ in water may act as a base and take a proton from water, and that is given by the equation:
So strictly speaking you only really need equation 3 to solve this problem, but the first two equations help you figure out how to solve it if you're not too familiar. What do I mean:
Equation 1: $\ce{HCN + H2O <=> H3O+ + CN-}$
As you correctly note, this is to remind you of the definition for $K_a$
Equation 2: $\ce{NaCN -> Na+ + CN-}$
Note that I've changed $\ce{<=>}$ to $\ce{->}$. This is because most sodium salts fully dissociate in aqueous solution and so this "equilibrium" doesn't exist and you can then consider all $\ce{NaCN}$ to be converted to $\ce{CN-}$ for your "initial" stage in the ICE table. Hence there's no need to write an ICE Table with respect to Equation 2.
Value of writing this out: In the case that you were given something sparingly soluble or did not fully dissociate, you would have to consider another equilibrium and this equation 2 would become significant. But in this case, $\ce{Na+}$ is a spectator ion which you can ignore.
Where does Equation 3 come from?
Equation 3 comes from the question itself (or rather what is happening in the system described). Remember, you've got a 1.0M solution of $\ce{NaCN}$. Since we've already established that it fully dissociates, at the "initial" stage of the system you have only $\ce{Na+}$ and $\ce{CN-}$ ions floating around. But from Equation 1 you know that $\ce{HCN}$ is a weak acid and so free $\ce{CN-}$ in water may act as a base and take a proton from water, and that is given by the equation:
First and Third (sic!) equations are necessary. The second one just says "when I say that 1M solution of NaCN was formed, understand it as 1M of Na+ and 1M of CN- were added to a system".
Your key equilibrium is [HA]=[H+][A-] and [HA]/[H+][A-]=Ka. You then say HA+A- = total_molarity, [HA]=X, [H+]=X, [A-] = total-molarity - X and
x/([x]*[ total-molarity - X] = Ka. You know Ka, total_molarity. You now need to find x. In case of NaCN HCN is HA, CN- is A-.
Just plug your concentrations in the formula and get the result. Quite boring, indeed.
First and Third (sic!) equations are necessary. The second one just says "when I say that 1M solution of NaCN was formed, understand it as 1M of Na+ and 1M of CN- were added to a system".
Your key equilibrium is [HA]=[H+][A-] and [HA]/[H+][A-]=Ka. You then say HA+A- = total_molarity, [HA]=X, [H+]=X, [A-] = total-molarity - X and
x/([x]*[ total-molarity - X] = Ka. You know Ka, total_molarity. You now need to find x. In case of NaCN HCN is HA, CN- is A-.
Just plug your concentrations in the formula and get the result. Quite boring, indeed.
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So strictly speaking you only really need equation 3 to solve this problem, but the first two equations help you figure out how to solve it if you're not too familiar. What do I mean:
As you correctly note, this is to remind you of the definition for $K_a$
Note that I've changed $\ce{<=>}$ to $\ce{->}$. This is because most sodium salts fully dissociate in aqueous solution and so this "equilibrium" doesn't exist and you can then consider all $\ce{NaCN}$ to be converted to $\ce{CN-}$ for your "initial" stage in the ICE table. Hence there's no need to write an ICE Table with respect to Equation 2.
Value of writing this out: In the case that you were given something sparingly soluble or did not fully dissociate, you would have to consider another equilibrium and this equation 2 would become significant. But in this case, $\ce{Na+}$ is a spectator ion which you can ignore.
Equation 3 comes from the question itself (or rather what is happening in the system described). Remember, you've got a 1.0M solution of $\ce{NaCN}$. Since we've already established that it fully dissociates, at the "initial" stage of the system you have only $\ce{Na+}$ and $\ce{CN-}$ ions floating around. But from Equation 1 you know that $\ce{HCN}$ is a weak acid and so free $\ce{CN-}$ in water may act as a base and take a proton from water, and that is given by the equation:
$$\ce{CN- + H2O <=> HCN + OH-}$$
So strictly speaking you only really need equation 3 to solve this problem, but the first two equations help you figure out how to solve it if you're not too familiar. What do I mean:
As you correctly note, this is to remind you of the definition for $K_a$
Note that I've changed $\ce{<=>}$ to $\ce{->}$. This is because most sodium salts fully dissociate in aqueous solution and so this "equilibrium" doesn't exist and you can then consider all $\ce{NaCN}$ to be converted to $\ce{CN-}$ for your "initial" stage in the ICE table. Hence there's no need to write an ICE Table with respect to Equation 2.
Value of writing this out: In the case that you were given something sparingly soluble or did not fully dissociate, you would have to consider another equilibrium and this equation 2 would become significant. But in this case, $\ce{Na+}$ is a spectator ion which you can ignore.
Equation 3 comes from the question itself (or rather what is happening in the system described). Remember, you've got a 1.0M solution of $\ce{NaCN}$. Since we've already established that it fully dissociates, at the "initial" stage of the system you have only $\ce{Na+}$ and $\ce{CN-}$ ions floating around. But from Equation 1 you know that $\ce{HCN}$ is a weak acid and so free $\ce{CN-}$ in water may act as a base and take a proton from water, and that is given by the equation:
$$\ce{CN- + H2O <=> HCN + OH-}$$
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