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 (OH-) and oxide (O2-) salts.  NaOH will provide one mole of OH- per mole of salt, but Ca(OH)2 will provide two moles of OH- per mole of salt.  Once you know the [H+] or [OH-], it is easy to change these into pH or pOH quantities.

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 OH-.

Example 1 - Finding the Ka of a weak acid from the pH of its solution.

A 0.10M solution of formic acid, HCOOH, has a pH = 2.38 at 25oC.  Calculate the Ka of formic acid. 1.  HCOOH (aq) D HCOO- + H+       Ka = [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)           [HCOO-](M)

Initial                0.10                             0.00                 0.00

Change             -x                                 +x                    +x

Equilibrium       0.10 -x                         x                      x

Ka   =   (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.

Ka = (x)(x)   = (4.2 x 10-3)2 = 1.8 x 10-4

(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 Ka of a weak acid.

Calculate the pH of a 0.10 M HF solution, given that the Ka = 6.8 x 10-4 at 25oC.

1.  HF(aq) D F- (aq) + H+(aq)             Ka = 6.8 x 10-4 = [F-]eq [H+]eq [HF]eq

2.  We know the initial [HF] and the Ka

3.                                 [HF] (M)                      [H+](M)           [F-](M)

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.

Ka = 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

Ka =  6.8 x 10-4 =  (x)(x)

(0.10 - x)

This needs to be expressed as a quadratic equation, and then solved  x2 + 6.8 x 10-4 x - 6.8 x 10-5 = 0           x = -b ± Ö b2 - 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 Ka 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 Ka 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 Ka.  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.

Ka =     (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 percent ionization is much higher in the dilute acid.  This is analogous to what we saw in the gaseous systems.  We saw that lowering the volume of a gaseous system shifted the equilibrium toward the side that had fewer moles of gas.  In aqueous systems, this same effect of concentration can be observed.  In acid dissociation problems, we have M2 in the products, and M in the reactants.  Decreasing the concentration of all of the species will make Q < Ka , since we are squaring the numerator (products), but not the denominator(reactant).  More of the acid will ionize, if the concentration of the acid is lower.

Example 4 Weak  base calculations.  These can be set up in exactly the same way, using Kb values and the equilibrium expression for weak bases.

Calculate the pH of a 0.15 M solution of NH3.  The Kb of ammonia at 25oC is 1.8 x 10-5.

1.  NH3 D NH4+ + OH-                       Kb = [NH4+]eq [OH-]eq [NH3]eq

2.  We know Kb and [NH3]i

3.                                 [NH3]               [NH4+]             [OH-]

Initial                0.15                 0.00                 0.00

Change             -x                     +x                    +x

Equilibrium       0.15 - x            x                      x

4.  We will assume that the OH- coming from the autoionization of water is very small and that very little of the ammonia will ionize.  We will check these assumptions when we finish the problem.

Kb = 1.8 x 10-5 = (x)(x)       x = 1.64 x 10-3 = [OH-]

(0.15)

1.64 x 10-3 x 100 = 1% = percent ionization and proof that we could make the simplifying assumption about x << 0.15.

0.15

This [OH-] is also much greater than 1 x 10-7 M OH- contributed by water.

The question asked for the pH of this solution.  We know the [OH-] so we can solve for pH two different ways.

pH + pOH = 14    or [H+] [OH-] = 1 x 10-14

[OH-] = 1.64 x 10-3 , 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 H2CO3.  This acid has two protons that can be dissociated, and two Ka values.

H2CO3 (aq) D H+ (aq) + HCO3- (aq)               Ka1 = 4.3 x 10-7

HCO3- (aq) D H+ (aq) + CO3- (aq)                  Ka2 = 5.6 x 10-11

The first acid, H2CO3 is a much stronger acid, judging from its higher value of Ka.  We will first calculate the pH and equilibrium concentrations after the first proton dissociates.

[H2CO3] (M)                [H+] (M)          [HCO3-] (M)

Initial                0.0037                         0.00                 0.00

Change             -x                                 +x                    +x

Equilibrium       0.0037 - x                    x                      x

4.3 x 10-7 =         x2          x = 4.0 x 10-5 (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 [H2CO3] = 0.0037 M, since the dissociation was so small.  [HCO3-] = [H+] = 4.0 x 10-5M after the first dissociation reaction.

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

HCO3- (aq) D H+ (aq) + CO32- (aq)                 Ka = 5.6 x 10-11

[HCO3-] (M)                [H+] (M)          [CO32-] (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 10-11  = [CO32-] (4.0 x 10-5 - y

To test our assumptions  5.6 x 10-11  x 100 = 1.4 x 10-4 %

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.