8.4 CALCULATIONS INVOLVING ACIDIC SOLUTIONS
CALCULATIONS INVOLVING STRONG ACID SOLUTIONS
When you are trying to understand the calculations involving strong acid solutions, one point should
come to mind. That is, strong acids ionize
in water! Therefore, the
concentration of hydrogen ions in the solution is equal to the given concentration of the acid in the
solution.
Example Problem 1: A 2.00 L hydrobromic acid solution (HBr(aq)), contains 0.070 mol of acid.
Calculate the concentration of hydrogen ions, hydroxide ions, pH and pOH.
nHNO3 = 0.070 mol, v = 2.00 L [HNO3] = 0.035 mol/L
Nitric Acid is a strong acid and ionizes 100% in water
HNO3(aq) ⇌ H+(aq) + NO31-(aq),
ratio between HNO3 and H+, therefore [H+] =
pH = -log [H+] =
Kw = [H+(aq)][OH-(aq)]
[OH-] = Kw
[H+]
pOH = -log[OH-]
CALCULATIONS INVOLVING WEAK ACID SOLUTIONS
When you are trying to understand the calculations involving weak acid solutions, you have to
remember that ionization is often incredibly small (at most, 50%) and there is often a large
concentration of acid molecules that haven’t ionized and a small concentration of protons and
anions. This is just like any equilibrium reaction and may require the use of an ICE table. All you
really is the Ka.
Another source of help is knowing the percentage ionization. For a weak acid: HA (aq) ⇌ H+(aq) + A-(aq)
Example Problem 2: Calculating pH from Ka
What is the hydrogen ion concentration and pH of a 6.18 x 10-3 mol/L solution of hydrocyanic acid,
HCN(aq)? The Ka for HCN (on Appendix B5) is 6.2 x 10-10.
Given: [HCN] = 6.18 x 10-3
Ka = 6.2 x 10-10
The ionization reaction is: HCN(aq) ⇌ H+(aq) + CN-(aq)
Set-Up an ICE Table for the ionization of HCN, where x is really the Δ[HCN] and write a Ka
expression
I
C
E
HCN(aq)
⇌
-3
6.18 x 10
-x
6.18 x 10-3 - x
H+(aq)
0
+x
+x
CN-(aq)
0
+x
+x
+
Check to see if we can simplify our expression with the
hundred rule: [HCN] = 6.18 x 10-3 = 9.97 x106
Ka
6.2 x 10-10
We can simplify the change in [HCN]
x = [H+]
pH = -log[H+]
pH = -log (
pH =
)
Example Problem 3: The Reverse – Calculating Ka from pH
A 0.050 mol/l solution of nicotinic acid, HC2H6NO2(aq, has a pH of 3.08. What is the Ka =?
[HC2H6NO2(aq)] = 0.050 mol/L
pH = 3.08
[H+] = 10-pH =
The ionization reaction is: HC2H6NO2(aq) ⇌ H+(aq) + C2H6NO2-(aq)
There is a 1:1 ratio between H+ and CH6NO2, therefore
[C2H6NO2-] =
Example Problem 4: Calculating Ka from Percentage Ionization
0.100 mol/L solution of hydrofluoric acid, HF(aq), has a percentage ionization of 7.8 %. Calculate
the Ka for HF.
[HF] = 0.100 mol/L
Percentage Ionizn = 7.8 %
[H+(aq)] = [% Ionizn][HF(aq)]
100
The ionization reaction is HF(aq) ⇌ H+(aq) + F-(aq)
I
C
E
HCN(aq)
0.100
-0.0078
0.092
⇌
H+(aq)
0
+0.0078
0.0078
+
CN-(aq)
0
+0.0078
0.0078
POLYPROTIC ACIDS AND THEIR IONIZATION
 A monoprotic acid is an acid that has
ionizable hydrogen
 A polyprotic acid is an acid has
ionizable hydrogen. H2SO4 is an example of a
acid and H3PO4 is an example of a
acid.
 Polyprotic acid ionize in two or more steps and ionize one hydrogen at a time. Each step
involving the ionizing of a hydrogen has its own Ka value. For a process that involves two
ionizable hydrogens, there may be a Ka1 and a Ka2
 Ka values will either be given to you or can be found on Table 4 on Page 522.
Example Problem 5: Calculating the pH of a Polyprotic acid solution
Calculate the pH of 0.10 mol/L solution of ascorbic acid (Vitamin C).
[H2C6H6O6] = 0.10 mol/L
Ka1 = 7.9 x 10-5
Ka2 = 1.6 x 10-12
Step 1: Write an ionization reaction for the first ionization. K a2 is smaller that Ka1. The first acid
will be the strongest.
H2C6H6O6(aq) ⇌ H+(aq) + HC6H6O6-(aq)
Step 2: Since this is a weak acid, prepare an ICE Table and Ka expression to determine the [H+]
7.9 x 10-5 = (x)(x)
0.10-x
7.9 x 10-5 = x2
0.10 –x
7.9 x 10-5 = x2
0.10
x2 = (0.10)(7.9 x10-5)
x2 = 7.9 x 10-6
x = √(7.9 x 10) = 2.8 x 10-3
[H+] = 2.8 x 10-3 mol/L
pH = -log[H+]
pH = -log (2.8 x10-3)
pH = 2.55
I
C
E
H2C6H6O6(aq)
0.10
-x
0.10 - x
⇌
H+(aq)
0
+x
x
+ HC6H6O6-(aq)
0
+x
x
Check to see if we can simplify our expression with the
hundred rule: [HC6H6O6] = 0.10 = 1.27 x104
Ka
7.9 x 10-5
We can simplify the change in [HC6H6O6], so
0.10-x ≈ 0.10
You can verify using the 5% Rule to check that your 100%
Rule Assumption was valid. If x/Ka is less than
5%, the such an assumption was valid
Step 3: Determine the equilibrium concentration of the other remaining entities.
[H2C6H6O6(aq)] = 0.10 –x
[HC6H6O6-] = x
= 0.10 – 2.8 x 10-3
[HC6H6O6-] = 2.8 x 10-3
= 0.097 mol/L
Step 4: Write an ionization reaction for the second ionization. Keep in mind that you already have
the concentration of the second acid from your previous ICE table. You can use all these quantities
to calculate the [C6H6O6-]
HC6H6O6-(aq) ⇌ H+(aq) + C6H6O6-(aq)
Ka2 =
HOMEWORK:
DO NOT COPY THIS PAGE THAT CONTAINS THE HOMEWORK:
Read Pages 512-524 and answer questions #1ac, 2a, 3a, 6, 7 on Page 525