**Acid/Base Calculations**

**Strong acids and Bases **

As we saw in the last lecture, calculations involving **strong acids** and **bases** are very straightforward. These species **dissociate completely** in water. So, [strong acid] = [H^{+}]. For strong bases, pay attention to the **formula**. Groups I and II both form hydroxide (^{-}^{2-}) salts.
NaOH will provide one mole of ^{-}_{2} will provide two moles of ^{-}^{+}] or [^{-}

**Weak acids and Bases**

These are more challenging, since they undergo **incomplete dissociation**, and exist as
systems at **equilibrium. **The best approach is the most systematic
approach.

1. Write the
balanced equation and an expression
for *K*.** **Compare these to the
information that you have been given, for example, initial concentrations,
equilibrium concentrations, the value of *K*,
etc. This should help you decide what
you need to find.

2. Define **x** as the unknown concentration, that
changes during the reaction. This
is determined by the stoichiometry of the reaction.

3. Construct a reaction table (ICE table) that incorporates the unknown.

4. **Make assumptions that will help to simplify
the calculations**. We haven't
discussed these yet, because they only apply to systems with **small equilibrium constants**, like weak
acids and bases.

a. Because the values of *K* are small, we know that very little of
the initial acid or base will be dissociated. In doing the preliminary
calculations, it is often safe to assume that the [HA]_{i} ≈
[HA]_{eq} . This should be
tested after the initial calculation, but it is very often true.

b. We will also assume that the [H^{+}]
and [OH^{-}] in a solution due to the **autoionization **of water is negligible. This only time this becomes important is
at very low (< 10^{-6} M) concentrations of acids or bases, when
water will be the main source of H^{+} and ^{-}

**Example 1** -
Finding the ** K_{a}** of a weak acid from the

A **0.10M** solution
of **formic acid**, HCOOH, has a **pH = 2.38** at 25^{o}C. Calculate the ** K_{a}** of formic
acid.

1. HCOOH (aq) D
HCOO^{-} + H^{+} *K*_{a} = [HCOO^{-}]_{eq}
[H^{+}]_{eq}

[HCOOH]_{eq}

We know the equilibrium concentration of H^{+}, since we were
given the pH of the solution. [H^{+}]_{eq}
= 10^{-2.38} = 4.2 x 10^{-3} M

2. We know the
initial concentration of HCOOH. We
will lose **x** moles of this as the
acid undergoes dissociation to form HCOO^{-} and H^{+}.

3. __[HCOOH](M)__ _{ }__[H ^{+}](M)__

Initial 0.10 0.00 0.00

Change -x +x +x

Equilibrium 0.10 -x x x

*K*_{a} = __(x)(x)__

(0.10 - x)

4. Assume that
all of the H^{+} comes from the acid, and none from water. We know that the [H^{+}]_{eq}
= 4.17 x 10^{-3} which is much higher than the 1.0 x 10^{-7} M
H^{+} from water. We can
also see that the concentration of HCOOH will change very little, from
0.10 to 0.10 - 4.17 x 10^{-3}. The change in concentration can be
ignored if it is less than **5%** of the
original concentration. 0.10 M x 5%
= 5.0 x 10^{-3}, so the change in [HCOOH] in this problem can be
ignored.

** K_{a}**
=

(0.10) (0.10)

We can also calculate the **percent ionization** for this problem. This will represent the relative number
of acid molecules which dissociate. It is calculated as :

**[H ^{+}]_{eq} x 100**

**
[HCOOH] _{i}**

In this case, the percent ionization = {(4.2 x 10^{-3})/
(0.10)}x 100 = **4.2%**

This indicates that very little of this acid dissociates
(ionizes) under these conditions.
For **strong **acids and bases,
the percent ionization is **100%**.

**Example 2** -
Finding the **pH** of a solution, given
the ** K_{a}
**of a weak acid.

Calculate the pH of a 0.10 M HF solution, given that the *K*_{a} = 6.8 x 10^{-4} at
25^{o}C.

1. HF(aq) D
F^{-} (aq) + H^{+}(aq) *K*_{a} = 6.8 x 10^{-4} =
[F^{-}]_{eq} [H^{+}]_{eq}__ __

[HF]_{eq}

2. We know the
initial [HF] and the *K*_{a}

3. __[HF]
(M)__ __[H ^{+}](M)__ [F

Initial 0.10 0.00 0.00

Change -x +x +x

Equilibrium 0.10 - x x x

4. We will again
assume that the [H^{+}] comes only from the acid, and that the decrease
in the initial concentration of HF is negligible.

*K*_{a} = 6.8 x 10^{-4} = __(x)(x) __ x
= 8.2 x 10^{-3}

(0.10)

Now we need to check if this is less than 5% of the original concentration, to see if our simplification is allowed.

__8.2 x 10__^{-3} x 100 = **8.2%**, so we **can't** use
the simplification!

0.10

*K*_{a} = 6.8 x 10^{-4} = __(x)(x)__

(0.10 - x)

This needs to be expressed as a **quadratic equation**, and then solved

x^{2} + 6.8 x 10^{-4}
x - 6.8 x 10^{-5} = 0 x
= -b ±
Ö
b^{2} - 4 a c

2 a

x = 7.9 x 10^{-3} M = [H^{+}]

pH = - log (7.9 x 10^{-3}) = 2.10

(If you hate solving the quadratic equation as much as I do,
there is an alternative, called **successive
iterations**. To use this method,
use the assumption that x is negligible compared to the initial concentration
of the acid, and solve for x as we did the first time, x = 8.2 x 10^{-3}. Now, **substitute this value into the equation**, by subtracting it from
[HF]. Multiply this answer by *K*_{a} and take its square root,
to get a new value for x. If you
did it right, it should be about 7.9 x 10^{-3}. Again, subtract this number from [HF].
multiply by *K*_{a} and take
the square root. You should again
get 7.9 x 10^{-3}, the right answer. At some point in this procedure, you
will always get a repeating value, the correct solution. To know if the 5% assumption is correct,
compare the [HA]_{i} to *K*_{a}. If it is at least 400 times bigger, the
assumption is O.K. If not, get our
your pencil and solve the quadratic, or use your calculator to converge on the
right answer.)

We can again calculate the **percent ionization** for this experiment. It will be {(7.9 x 10^{-3})/(9.10)}x
100 = **7.9 %**

So, a 0.10 M solution of HF will ionize to a greater extent (7.9%) than a 0.10 M solution of formic acid, HCOOH (4.2%). It is a stronger acid.

**Example 3 ** What is the pH of a **0.010 M** solution of HF? This is the same system as in
Example 2, just with a more dilute concentration of HF. So, all that will change will be the
initial concentration of HF.

*K*_{a}
= __(x)(x)__ = 6.8 x 10^{-4}

(0.010 - x)

Solving for x (quadratic equation or successive iterations)
gives x = [H^{+}] = 2.3 x 10^{-3} M

So, the pH = 2.64. The pH is slightly less acidic,
because the concentration of H^{+} in the more dilute system is
lower.

The **percent
ionization** = ([H^{+}] /
[HF]) x 100 = {(2.3 x 10^{-3})/ (0.010)}x 100 = **23.0%**

Although the **[H ^{+}]
is lower**, the

**Example 4** **Weak
base calculations**. These
can be set up in exactly the same way, using *K*_{b} values and the equilibrium expression for weak
bases.

Calculate the pH of a 0.15 M solution of NH_{3}. The *K*_{b}
of ammonia at 25^{o}C is 1.8 x 10^{-5}.

1. NH_{3}
D
NH_{4}^{+} + OH^{-} *K*_{b} = [NH_{4}^{+}]_{eq}
[^{-}_{eq}

[NH_{3}]_{eq}

2. We know *K*_{b} and [NH_{3}]_{i}

3. [NH_{3}] [NH_{4}^{+}] [^{-}

Initial 0.15 0.00 0.00

Change -x +x +x

Equilibrium 0.15 - x x x

4. We will
assume that the ^{-}

*K*_{b} =
1.8 x 10^{-5} = __(x)(x)__ x = 1.64 x
10^{-3} = [^{-}

(0.15)

__1.64 x 10 ^{-3}__ x 100 =

0.15

This [^{-}^{-7} M OH^{-} contributed by
water.

The question asked for the **pH** of this solution. We
know the [^{-}

pH + pOH = 14 or [H^{+}] [^{-}^{-14}

[^{-}^{-3} , so **pOH** =
-log(1.64 x 10^{-3}) = 2.79

2.79 + pH = 14 pH = 11.21

or [H^{+}]
[1.64 x 10^{-3}] = 1 x 10^{-14} [H^{+}] = 6.10 x 10^{-12} pH = 11.21.

**Example 5 Polyprotic
acids**

Find the pH of a 0.0037 M solution of H_{2}CO_{3}.
This acid has two protons that can
be dissociated, and two *K*_{a}
values.

H_{2}CO_{3} (aq) D
H^{+} (aq) + HCO_{3}^{-} (aq) *K*_{a1} = 4.3 x 10^{-7}

HCO_{3}^{-} (aq) D
H^{+} (aq) + CO_{3}^{-} (aq) *K*_{a2} = 5.6 x 10^{-11}

The first acid, H_{2}CO_{3} is a much
stronger acid, judging from its higher value of *K*_{a}. We will
first calculate the pH and equilibrium concentrations after the first proton
dissociates.

[H_{2}CO_{3}]
(M) [H^{+}]
(M) [HCO_{3}^{-}]
(M)

Initial 0.0037 0.00 0.00

Change -x +x +x

Equilibrium 0.0037 - x x x

4.3 x 10^{-7} =
__ x ^{2} __ x = 4.0 x 10

(0.0037 - x)

To test the assumption that x is <<< 0.0037, __4.0 x 10__^{-5} x 100
= 1.1%, so the assumption is O.K.

3.7
x 10^{-3}

So, the [H_{2}CO_{3}] = 0.0037 M, since the
dissociation was so small. [HCO_{3}^{-}]
= [H^{+}] = 4.0 x 10^{-5}M after the first dissociation
reaction.

We can now use these concentrations as the **initial concentrations** in the second
acidic reaction.

HCO_{3}^{-} (aq) D
H^{+} (aq) + CO_{3}^{2-} (aq) *K*_{a} = 5.6 x 10^{-11}

[HCO_{3}^{-}]
(M) [H^{+}]
(M) [CO_{3}^{2-}]
(M)

Initial 4.0
x 10^{-5} 4.0
x 10^{-5} 0.00

Change -y +y +y

Equilibrium 4.0
x 10^{-5} - y 4.0
x 10^{-5} + y y

5.6 x 10^{-11} = __(4.0 x 10 ^{-5} + y)(y)__ y = 5.6 x 10

(4.0 x 10^{-5} - y

To test our assumptions
__5.6 x 10 ^{-11}__

4.0 x 10^{-5}

So, the pH at equilibrium will be determined exclusively by
the first step, [H^{=}] = 4.0 x 10^{-5}, pH = 4.4.