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Strong Acids and Bases

Skills to develop

Strong Acids and Bases

The animation here shows the formation of H3+O ions and OH- ions in an aqueous system.

Acids and bases that are completely ionized when dissolved in water are called strong acids and strong bases There are only a few strong acids and bases, and everyone should know their names and properties. These acids are often used in industry and everyday life.

The concentrations of acids and bases are often expressed in terms of pH, and as an educated person, you should have the skill to convert concentrations into pH and pOH. The pH is an indication of the hydrogen ion concentration, [H+].

Strong Acids

Strong acids
TypeFormula
Hydrogen
halides		HCl HBr HI
Oxyacids
of halogens, 		HClO3
HClO4 		HBrO3
HBrO4 		HIO3
HIO4
Sulfuric acidH2SO4
Nitric acidHNO3
Strong acids are acids that are completely or nearly 100% ionized in their solutions. Here are some common strong acids: Ionization of a strong acid HA can be represented by: HA = H+ + A-
          x       x where x is the concentration of H+, [H+]. For a strong acid, [H+] = [A-] = concentration of acid (= x), if x is much greater than 1x10-7 (represented as e-7). For a very dilute strong acid solution with concentration less than 1E-7, the pH is dominated by the autoionization of water, H2O = H+ + OH-,       Kw = 1e-14 at 298 K.

The pH and pOH Scales

The pH and pOH
scale at 298 K
pH[H+]pOH
1 0.1 13
2 0.01 12
3 0.00111
4 1e-4 10
5 1e-5 9
6 1e-6 8
The pH scale is defined as the negative log of the concentration of H+: pH = -log[H+] The pOH scale is defined as the negative log of the concentration of OH-, [OH-]: pOH = -log[OH-] With this scale, calculating the pOH can be done in the same manner as the pH scale.

Example 1

Calculate the pH of a solution with 1.2345E-4 M HCl

Solution
The solution of a strong acid is completely ionized. Thus, [H+] = 1.234e-4.

pH = -log(1.234e-4) = 3.909

Discussion
What is the pH for a solution containing 1.234 M [HCl]? pH = 0.0913

Example 2

Calculate the pH of a stock HCl solution that is 32% by mass HCl.

Solution
The density of such a solution is needed before we can calculate the pH. Since the density is not on the label, we need to find it from the Material Safety Data Sheet, which gives the specific gravity of 1.150. Thus, the amount of acid in 1.0 L is 1150 g.

The amount of HCl = 1000*1.150*0.32
      = 368 g (1 mol/36.5 g <- molar mass of HCl)
      = 10.08 M
      = [H+]

pH = -log(10..08) = -1.003

Discussion
Yes, pH have negative values if [H+] > 1.0

Check out the information on nitric acid, and calculate the pH of a stock nitric acid solution.

Example 3

Calculate the pH of a solution containing 1.00E-7 of HCl.

Solution
[H+] = 1.0e-7 M from the strong acid, and if x is the amount from the ionization of water, then we have the equilibrium due to the autoionization of water: 

  HCl  =  H+  +  Cl-
         1E-7   1E-7     <--- [H+] from the acid

  H2O  =  H+  +  OH-
      (1E-7)+x    x      <--- we don't know yet
Recall that Kw = [H+] [OH-] = 1E-14, due to the ionization equilibrium of water in the solution: {(1.00E-7)+x} x = 1E-14
x2 + 1.00e-7x - 1.00E-14 = 0 Solving this equation for x results in x = {-1.00E-7 + (1.00E-14 + 4*1.00E-14)1/2} / 2
      = 0.61E-7
[H+] = (1.00 + 0.61)E-7 M
pH = -log(1.61E-7) = 6.79

Discussion
If you require only 1 significant figure, the pH is about 7.

Strong Bases

Strong bases are completely ionized in solution. For example, KOH dissolve in water in the reaction KOH = K+ + OH-.

Strong bases
NameFormula
Sodium
hydroxide		NaOH
Potassium
hydroxide		KOH
Cesium
hydroxide		CsOH
Calcium
hydroxide		Ca(OH)2
Relative to strong acids, there are fewer number of strong bases. Most strong bases are alkali hydroxides. Calcium oxide is considered a strong base, because it is completely, almost completely, ionized. However, the solubility of calcium hydroxide is very low. When Ca(OH)2 dissolve in water, the ionization reaction is as follows:

Ca(OH)2 = Ca2+ + 2 OH-. The concentration of OH- is twice the concentration of Ca2+, [OH-] = 2 [Ca2+]

Example 4

Calculate the pOH of a solution containing 1.2345E-4 M Ca(OH)2.

Solution
Based on the ionization,

[OH-] = 2*1.234e4 = 2.468E-4 M
pOH = -log(2.468E-4)
  = 3.61

Discussion
The molar solubility of calcium hydroxide is 0.013 M Ca(OH)2. Calculate the pOH. pOH = 1.58

Autoionization of Water

The equilibrium product Kw = [H+] [OH-] is a constant at a definite temperature due to the autoionization of water, H2O = H+ + OH-. At 298 K, Kw = 10-14 and the following relationship in any aqueous solution is obvious, pOH + pH = 14 at 298 K. Of course, when T is higher than 298 K, pH + pOH is slightly less than 14 due to a higher degree of ionization of water. Conversely, at low temperatures, pH + pOH is larger than 14 due to less degree of ionization.

Confidence Building Questions

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