Basic Acid-base Review Problems
The problems below start with strong acids and bases and then moves to weak acids and bases.
Note that the approximations that avoid the quadratic formula are possible if the pH is not near
7.0. The problems employ the 5% approximation, which says that you may ignore [H+] and
[OH-] from water if they are less the 5% of the final values.
You should try working the problems using the quadratic formula to prove that the answers are
the same.
1. What is the pH of a 0.0115 M HCl solution?
The only two materials in the solution are water and HCl. The water alone would
provide 10-7 Molar hydrogen ion. The HCl is a strong and soluble acid, so all of the
acid is in the form of ions, providing 0.0115 Molar hydrogen ion. There is no need to
consider the hydrogen ion from water as it is much less than five percent of the total
hydrogen ion concentration. The concentration of hydrogen ion is adequately
estimated at 0.0115 M.
𝑝𝑝𝑝𝑝 = −𝑙𝑙𝑙𝑙𝑙𝑙10 [𝐻𝐻+]
𝑝𝑝𝑝𝑝 = −𝑙𝑙𝑙𝑙𝑙𝑙10 (0.0115) = 1.9
2. Find the pH of 0.0815 M NaOH solution.
The only two materials in the solution are water and NaOH. The water alone would
provide 10-7 Molar hydroxide ion. The NaOH is a strong and soluble base, so all of
the base is in the form of ions, providing 0.0815 Molar hydroxide ion. The hydroxide
ion is most important in this problem. The hydroxide ion from the dissociation of
water is much less than five percent of the hydroxide ion from sodium hydroxide, so
the concentration of hydroxide ion is 0.0815 M.
𝑝𝑝𝑝𝑝𝑝𝑝 = −𝑙𝑙𝑙𝑙𝑙𝑙10 [𝑂𝑂𝑂𝑂−]
𝑝𝑝𝑝𝑝𝑝𝑝 = −𝑙𝑙𝑙𝑙𝑙𝑙10 (0.0815) = 1.1
𝑝𝑝𝑝𝑝 = 14 − 𝑝𝑝𝑝𝑝𝑝𝑝 = 14 − 1.1 = 12.9
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3. Find the pH of 0.00372 M Ba(OH)2 solution.
Ba(OH)2 is a strong base, that is, up to its solubility limits the compound will be
completely ionized in water solution. [(OH)-] is twice the [Ba(OH)2] because for each
mol of Ba(OH)2 there are two mols of hydroxide ion. The ionization of water, the only
other possible source of hydroxide ions, produces nowhere near five percent of the
hydroxide ion available from Ba(OH)2, and so:
[(OH)-] = 2 [Ba(OH)2] = 2 (0.00372 M) = 0.00744 M
𝑝𝑝𝑝𝑝𝑝𝑝 = −𝑙𝑙𝑙𝑙𝑙𝑙10 (0.00744) = 2.1
𝑝𝑝𝑝𝑝 = 14 − 𝑝𝑝𝑝𝑝𝑝𝑝 = 14 − 1.1 = 11.9
4. Find the pH of 0.12 M HC2H3O2 (Acetic acid)
The only two things in this solution are water and acetic acid, a weak acid. The Ka of
acetic acid is 1.74 x 10-5. Water and acetic acid could contribute hydrogen ion to this
mixture. If water were the only possible contributor, it would give 10-7 M hydrogen
ion. If the acetic acid were the only contributor, we could calculate it from the
equilibrium expression.
Note that this is the simplified approach that avoids the quadratic formula. It only
works for cases like this where the dissociation involves a monoprotic weak acid.
When in doubt, use the quadratic formula!
Since the hydrogen ion from the acetic acid would be exactly the same concentration
as the acetate ion, [H+] = [(C2H3O2)-]. And we can simplify by substituting [H+]2 for
[H+] [(C2H3O2)-].
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Now we can substitute our numbers for the Ka and [HC2H3O2] to find that [H+] = the
square root of (0.12 x 1.74 E-5). The [H+] due to acetic acid is 1.4450 E-3 M.
The [H+] from the acetic acid is less than five percent of the [acetic acid], so there is
no need to get a more accurate answer by subtracting the [H+] from the [acetic acid].
The pH is 2.84, rounded to 2.8.
5. pH of 0.255 M NH4OH
The Kb of NH4OH is 1.78 x 10-5.
The ionization equilibrium expression for NH4OH or any other weak base is similar to
the ionization equilibrium expression for a weak acid, but this time the equation will
be solved for the hydroxide ion concentration.
The hydroxide ion is in excess. The two sources of hydroxide ion are the [H+] and
water, but the contribution from the water is negligible.
The other assumption, that the amount of dissociated ion is negligible, is also valid
because the
concentration of base is
so large.
(Try it if you don't
believe it.)
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Now substitute the numbers and do the math.
𝑝𝑝𝑝𝑝𝑝𝑝 = −𝑙𝑙𝑙𝑙𝑙𝑙10 (0.00213) = 2.67
𝑝𝑝𝑝𝑝 = 14 − 𝑝𝑝𝑝𝑝𝑝𝑝 = 14 − 2.7 = 11.3
6. Calculate the pH of 0.578 M H3PO4
In a solution that has only water and phosphoric acid there are four possible sources of
hydrogen ion, the water and the three ionizations of the phosphoric acid. Each
ionization has a different Ka.
H3PO4 (H2PO4)- + H+
(H2PO4)- (HPO4)2- + H+
(HPO4)2- (PO4)3- + H+
first ionization
second ionizaton
third ionization
Ka = 6.92 x10-3
Ka = 6.17 x10-8
Ka = 2.09 x10-12
It is clear that the first ionization of phosphoric acid at this concentration is much
greater than the third ionization of phosphoric acid, but the big questions are whether
we need to consider the second ionization of phosphoric acid or the ionization of
water.
The hydrogen ion concentration from the first ionization can be calculated by the
usual square root of ([acid] Ka) to get an initial estimate.
From the initial estimate, the [H+] from the first ionization (0.063243656 M) will be
more than 5% of the acid concentration (0.578 M). We will have to use the more
accurate form of the equation that includes the decrease in unionized acid
concentration by the ionized forms.
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Will we have to also account for the second ionization? The number we found as first
estimate for the [H+] can also serve as estimate of the [H2PO4)-], the anion in the first
ionization. But the anion of the first ionization is the original acid in the second
ionization and the estimate of the [H+] can also be used in the second ionization
equation. If we solve for the anion of the second ionization equation, we would get an
estimate of the hydrogen ion that the second equation would produce.
This is an interesting result. Notice the [H+] from the second ionization is very close to
the second kA when there is nothing but water and phosphoric acid in the solution.
What would happen at other concentrations of phosphoric acid? At other pH's whith
the same initial concentration of phosphoric acid?
The concentration estimate of the hydrogen ion from the second ionization is far
below 5% of the hydrogen ion from the first ionization, so the second ionization does
not need to be considered. But we do need to consider the decrease in acid
concentration from ionization of the first reaction, so we use the quadratic form of the
equation substituted into the quadratic equation.
When you substitute the numbers, the math looks like:
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As you can see when you do the math, the [H+] is 0.059878231, more than 5%
different from the original estimate.
𝑝𝑝𝑝𝑝 = −𝑙𝑙𝑙𝑙𝑙𝑙10 [𝐻𝐻+]
𝑝𝑝𝑝𝑝 = −𝑙𝑙𝑙𝑙𝑙𝑙10 (0.0598) = 2.2
7. What is the pH of 0.16 M HCl and 0.072 M phosphoric acid?
There are five sources of hydrogen ion in this solution. Count them. Water, HCl, and
the three ionizations of the phosphoric acid. Likely the water and the lesser two of the
three ionizations of phosphoric acid will not significantly contribute to the hydrogen
ion concentration.
Since the HCl is a strong acid, it contributes 0.16 Molar hydrogen ion to the solution,
but the phosphoric acid is a weak acid and the [H+] from it cannot calculated by the
square root of (kA [H3PO4]) because the [H+] is not equal to the [H2PO4]. We must go
back to the ionization equilibrium equation for the first ionization of phosphoric acid.
We can look up the Ka. The [H+] is the same as the [HCl], and the concentration of
phosphoric acid is given. The [H2PO4] will be equal to the concentration of hydrogen
ion contributed by the first ionization of the phosphoric acid.
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So the pH depends only on the [HCl].
𝑝𝑝𝑝𝑝 = −𝑙𝑙𝑙𝑙𝑙𝑙10 [𝐻𝐻+]
𝑝𝑝𝑝𝑝 = −𝑙𝑙𝑙𝑙𝑙𝑙10 (0.16) = 0.8
8. Find the pH of 1.25 M acetic acid and 0.75 M potassium acetate.
Acetic acid kA = 1.74 E-5
pKA = 4.76.
This is a genuine buffer problem. Added to the water are a weak acid and a salt
containing the anion of the acid.
There are two good ways to work buffer problems, with the Henderson- Hasselbach
equation or with the ionization equilibrium expression of the weak acid or base. I
personally have a mental block against the H-H equation because I can never
remember whether it uses a positive or negative log and which concentration goes on
top. You can use it if you wish. Particularly if you need to calculate buffers often, you
should engrave it upon your gray matter. If you really need it and can't remember it,
you can derive it from the ionization equlilbrium expression.
There are three cautions you need to observe with either equation: (1) Make sure you
are using the correct concentration for each variable, (2) check to see if the numbers
you propose to use are going to be within the 5% rule for simplification, and (3)
estimate the answer from what you know and make sure your final answer is
reasonable.
Before actually doing the problem, estimate the answer from your own reasoning. In
this case, the pKa of acetic acid is 4.76. The rule is that an equimolar buffer has a pH
equal to the pKa and in this problem there is less potassium acetate than acetic acid,
so the pH must be lower (more acid) than the pKa within a pH unit or so. If the acetic
acid were the only solute, the pH estimate would be the square root of (acid
concentration times Ka).
The answer should be somewhere between pH of 2.3 and 4.8
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The majority of the acetate ion will be from the potassium acetate. Is it right that the
total acetate ion concentration will be equal to the concentration of the potassium
acetate? Or will the acetate ion concentration from the ionization of the acetic acid
contribute more than 5%? The potassium acetate concentration is 0.75 M. The acetate
ion concentration from acetic acid would be 0.00466 M, less than 5% of 0.75 M even
without the common ion effect. We can safely use 0.75 as the concentration of acetate
ion.
Will the concentration of unionized acid be a problem? The measured concentration is
1.25 M and the ionized amount is 0.00466 M, far less than 5% of 1.25 M.
As threatened, we can use the ionization equilibrium expression of acetic acid for the
main equation for this problem, substituting for the kA, substituting the concentration
of potassium acetate for the concentration of acetate, substituting the concentration of
acetic acid, and solving for the hydrogen ion concentration to get the pH.
The answer of pH = 4.5 is a reasonable one by our estimation because it is more acid
than the pKa of 4.76.
It is a little easier to do this problem by the Henderson- Hasselbach equation, if you
are sure you know it. You must still make sure you are substituting correctly and that
your assumptions for simplification are valid (within 5%). The H-H equation is not
much good for solutions in which either the acid or ion concentrations are more than
ten times one another or in which the concentration of either material is less than one
hundred times the kA because it doesn't easily adapt to a quadratic form.
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