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To find the pH of a 1.0 M solution of $\ce{NaCN}$, given $\ce{Ka(HCN) =4.9E(-10)}$

The solution (refer end of question) uses three equations:

Equation 1 : $\ce{HCN + H2O <-> CN- + H+}$

Equation 2: $\ce{NaCN <-> CN- + Na+}$

Equation 3: $\ce{CN- + H2O <-> HCN + OH-}$

I am confused as to:

1) the logic behind why we need these three equations (up to this point in the course, all the similar questions have been of the form e.g. find pH of NH3 given Kb(NH3), and so we only used the NH3 acid base reaction equation.

I assume equation 1 is necessary because the Ka is given for HCN and Equation 2 is necessary because we are asked to find the pH of NaCN. Where does Equation 3 come from? We did talk about spectator ions in class so does it have something to do with Na being a spectator ion?

2) why the ICE (intial, change, at equillibrium) table to find x is written in terms of Equation 3 and not, for example, Equation 2.

The solution from the textbook is given belowTextbook Solution:

Textbook Solution

Thank you.

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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:

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:

$$\ce{CN- + H2O <=> HCN + OH-}$$

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  • $\begingroup$ Great. Thank you for explaining that IT Tsoi :) $\endgroup$ – K-Feldspar Jul 14 '16 at 23:58
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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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