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Some weak acids and weak bases ionize to such an extent that the simplifying assumption that x is small relative to the initial concentration of the acid or base is inappropriate. As we solve for the equilibrium concentrations in such cases, we will see that we cannot neglect the change in the initial concentration of the acid or base, and we must solve the equilibrium equations by using the quadratic equation.

Equilibrium concentrations in a solution of a weak acid

Sodium bisulfate, NaHSO 4 , is used in some household cleansers because it contains the HSO 4 ion, a weak acid. What is the pH of a 0.50- M solution of HSO 4 ?

HSO 4 ( a q ) + H 2 O ( l ) H 3 O + ( a q ) + SO 4 2− ( a q ) K a = 1.2 × 10 −2

Solution

We need to determine the equilibrium concentration of the hydronium ion that results from the ionization of HSO 4 so that we can use [ H 3 O + ] to determine the pH. As in the previous examples, we can approach the solution by the following steps:

Three rectangles are shown with right pointing arrows between them. The first is labeled “Determine x and the equilibrium concentrations.” The second is labeled “Solve for x and the equilibrium concentrations. The third is labeled “Check the math.”
  1. Determine x and equilibrium concentrations . This table shows the changes and concentrations:
    This table has two main columns and four rows. The first row for the first column does not have a heading and then has the following in the first column: Initial concentration ( M ), Change ( M ), Equilibrium ( M ). The second column has the header of “H S O subscript 4 superscript negative sign plus sign H subscript 2 O equilibrium sign H subscript 3 O superscript positive sign plus sign S O subscript 4 superscript 2 superscript negative sign.” Under the second column is a subgroup of four columns and three rows. The first column has the following: 0.50, negative x, 0.50 plus sign negative x equals 0.50 minus x. The second column is blank for all three rows. The third column has the following: approximately 0, x, 0 plus sign x equals x. The fourth column has the following: 0, x, 0 plus sign x equals x.
  2. Solve for x and the concentrations . As we begin solving for x , we will find this is more complicated than in previous examples. As we discuss these complications we should not lose track of the fact that it is still the purpose of this step to determine the value of x .
    At equilibrium:
    K a = 1.2 × 10 −2 = [ H 3 O + ] [ SO 4 2− ] [ HSO 4 ] = ( x ) ( x ) 0.50 x

    If we assume that x is small and approximate (0.50 − x ) as 0.50, we find:
    x = 7.7 × 10 −2

    When we check the assumption, we calculate:
    x [ HSO 4 ] i

    x 0.50 = 7.7 × 10 −2 0.50 = 0.15 ( 15 % )

    The value of x is not less than 5% of 0.50, so the assumption is not valid. We need the quadratic formula to find x .
    The equation:
    K a = 1.2 × 10 −2 = ( x ) ( x ) 0.50 x

    gives
    6.0 × 10 −3 1.2 × 10 −2 x = x 2+

    or
    x 2+ + 1.2 × 10 −2 x 6.0 × 10 −3 = 0

    This equation can be solved using the quadratic formula. For an equation of the form
    a x 2+ + b x + c = 0 ,

    x is given by the equation:
    x = b ± b 2+ 4 a c 2 a

    In this problem, a = 1, b = 1.2 × 10 −3 , and c = −6.0 × 10 −3 .
    Solving for x gives a negative root (which cannot be correct since concentration cannot be negative) and a positive root:
    x = 7.2 × 10 −2

    Now determine the hydronium ion concentration and the pH:
    [ H 3 O + ] = ~ 0 + x = 0 + 7.2 × 10 −2 M

    = 7.2 × 10 −2 M

    The pH of this solution is:
    pH = −log [ H 3 O + ] = −log 7.2 × 10 −2 = 1.14

Check your learning

(a) Show that the quadratic formula gives x = 7.2 × 10 −2 .

(b) Calculate the pH in a 0.010- M solution of caffeine, a weak base:

C 8 H 10 N 4 O 2 ( a q ) + H 2 O ( l ) C 8 H 10 N 4 O 2 H + ( a q ) + OH ( a q ) K b = 2.5 × 10 −4

(Hint: It will be necessary to convert [OH ] to [ H 3 O + ] or pOH to pH toward the end of the calculation.)

Answer:

pH 11.16

The relative strengths of strong acids and bases

Strong acids, such as HCl, HBr, and HI, all exhibit the same strength in water. The water molecule is such a strong base compared to the conjugate bases Cl , Br , and I that ionization of these strong acids is essentially complete in aqueous solutions. In solvents less basic than water, we find HCl, HBr, and HI differ markedly in their tendency to give up a proton to the solvent. For example, when dissolved in ethanol (a weaker base than water), the extent of ionization increases in the order HCl<HBr<HI, and so HI is demonstrated to be the strongest of these acids. The inability to discern differences in strength among strong acids dissolved in water is known as the leveling effect of water    .

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Source:  OpenStax, Chemistry. OpenStax CNX. May 20, 2015 Download for free at http://legacy.cnx.org/content/col11760/1.9
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