Limiting Reactant
Consider what happens when we mix two solutions, potassium chloride and lead (II) nitrate
What could happen?
When two ionic solutions mix, the cations and anions of one solution could interact with the cation and
anion from the other solution. This can result in no change. All of the ions could just remain in solution.
But, the affinity of one ion for another may also result in the formation of precipitates.
Consider the ions:
So what are the other possibilities?
According to general solubility rules, nitrates and group (I) metal salts are soluble. So, combining
potassium and nitrate ions doesn’t do anything interesting. But lead(II) chloride is not soluble. A white
precipitate forms. Then the overall reaction is:
Balancing this reaction, we get:
As a net ionic equation, what occurs is clearer. Two ions combine and fall out of solution:
Consider the following situation:
In lab, you are given 0.15 M 𝐾𝐶𝑙 and 0.10 M𝑃𝑏 𝑁𝑂 . You combine 10 ml of each solution.
Decanting and drying the precipitate, how much mass will you find (presuming no loss in the
decanting process).
What do we do to solve this problem? We want to know how many grams of lead (II) chloride are
produced. But we need to check both reactants to see how much precipitate they could possibly
produce.
Note that the third factor is found by looking at the balanced reaction. The coefficients in the balanced
reaction give mole rations between products and reactants. The last factor is just the molar mass as
calculated from a periodic table.
Now, repeat the process with the lead (II) nitrate:
While there is enough lead (II) nitrate to make 0.28 g of precipitate, there is only enough potassium
chloride to make 0.21 g of precipitate. So, we expect to see 0.21 g of precipitate. There will be chloride
ions remaining in solution (as well as the spectator potassium and nitrate ions). There are not enough
lead ions to latch onto all of the chlorines and remove them from solution.
This style of problem is called a limiting reactant problem. One of the reactants sets and upper bound on
the amount of product that can be produced and some of the other reactant is left over.
Now, what if you actually measure 0.19 g of precipitate? What is the yield of the reaction? Theoretically,
you should get 0.21 g. So, as a ratio, expressed as a percent, we have:
Side note:
Recall that Arrhenius definition of acids and bases is that an acid gives
and a base gives
example, we could consider the reaction of hydrochloric acid and sodium hydroxide:
donates a proton and
. For
donates an hydroxide.
We could also write just the net ionic equation:
Brønsted and Lowery defined acids as proton donors and bases as proton acceptors. For instance,
consider the reaction of hydrochloric acid with ammonia:
donates a proton and
accepts it.
As a net ionic equation, we would write:
Lewis makes the broadest definition of acids and bases. An acid is an electron pair acceptor and a base is
an electron pair donor. If we consider the first two reactions,
, which has no electrons of its own
latches onto the pair of electrons offered by the
ion or the non-bonding pair on the nitrogen in
ammonia. As a result, a covalent bond is formed. We call this bond a coordinate covalent bond because
both electrons come from a single source.
But it also includes reactions that would not be considered acid-base reactions under the other two
definitions. For instances, consider boron trifluoride reacting with sodium fluoride.
The real heart of this reaction, though, is seen in the net ionic equation:
Consider the Lewis structures of
and
F
F
F
B
F
The fluoride ion donates a pair to the boron which, with three valance electrons, normally only forms an
sp2 hybrid – half-filling each orbital so that it can form three bonds with three elements which can
donate a single electron, like the fluorines. But, the fluoride ion, has a lone pair (well four of them) that
is can donate. So, this allows boron to form a covalent bond with the fluoride ion, using a pair of
electrons solely from the fluoride ion.
The chloride ion in the lead (II) chloride precipitation is donating a pair of electrons to the lead (II) ion.
So, in a broad sense, the precipitation reaction shares a great deal with a Lewis acid-base reaction. Of
course, the bond is ionic in character, rather than covalent. So it is not – by strict definition – the
coordinate covalent bond formed in a traditional Lewis acid-base reaction.