December 11, 2009

Dissociation of Weak Acid and Base Problems

Some problems are solved using both approximation and the quadratic equation. Values in the parenthesis are obtained using the quadratic equation.

Approximation is made by assuming that the amount of dissociated weak acid or base is negligible such that the given amount of undissociated acid or base remains practically the same.

When approximation introduces large % error, the quadratic equation alone is used as indicated in the Answer part of the problem.

For convenience, problems are grouped into aqueous solutions containing:
  • a weak acid or a weak base
  • a strong acid or a strong base
  • a salt of a weak acid or a salt of a weak base
  • a weak acid and its salt (weak acid/conjugate base composition) or a weak base and its salt (weak base/conjugate acid composition)
    • For a more detailed discussion of this topic and other chemistry topics, go to www.chemistrypartner.blogspot.com.

    Aqueous Solution of a Weak Acid or a Weak Base

    The pH of an aqueous solution containing a weak acid or a weak base is calculated from their dissociation constant equations shown below:
    1.If a solution of HCOOH has a concentration of 0.49 M, what is its pH? Ka for formic acid is 1.8 x 10-4.
    Answer: pH = 2.03 ( 2.03 )
    2.What is the pH of an aqueous solution of CH3(CH2)2CH2NH2 having a concentration of 0.04 M? The dissociation constant of 1-butylamine is 4 x 10-4.
    Answer: pH = 11.58 (Quadratic equation is used.)
    3.What is the pH of a 0.85 M HNO2 solution. ( Ka = 4.5 x 10-4 )
    Answer: pH = 1.71; ( 1.71)
    4.What is the pH of an aqueous solution of (CH3)2NH having a concentration of 0.08 M? The dissociation constant of dimethylamine is 5.9 x 10-4.
    Answer: pH = 11.84 ( 11.82)
    5.What is the pH of a 0.13 M HC2H2ClO2 solution. ( Ka = 1.36 x 10-3 )
    Answer: pH = 1.9 (Quadratic equation is used.)

    Aqueous Solution of a Strong Acid or a Strong Base

    The [H3O+] or pH of an aqueous solution containing a strong acid or base is calculated using the ion-product constant of water, Kw.

    Kw = [H3O+] [OH-]

    1 x 10-14 = [H3O+] [OH-]

    6.Find the [H3O+] and [OH-] of a 0.48 M of HCN solution. Ka = 6.2 x 10-10.
    Answer: [H3O+] = 1.73 x 10-5 M; [OH-] = 5.8 x 10-10 M ( 1.73 x 10-5; 5.8 x 10-10 )
    7.A one liter aqueous solution contains 3.37 g KOH. What is the pH of this solution? The dissociation constant for water, Kw, is 1 x 10-14. ( KOH = 56.1056 g )
    Answer: pH = 12.78
    8.A one liter aqueous solution contains 6 g NaOH. What is the pH of this solution? The dissociation constant for water, Kw, is 1 x 10-14. ( NaOH = 39.99707 g )
    Answer: pH = 13.18

    Aqueous Solution of a Salt of a Weak Acid or a Salt of a Weak Base

    The general relationship of the conjugate acid/base pairs is used to solve problems 9 to 14.

    This relationship is expressed as follows:

    KaKb = Kw

    KaKb = 1 x 10-14

    9.Calculate the pH of a 0.71 M NaC6H5COO solution. (Ka for benzoic acid is 6.3 x 10-5; Kw = 1 x 10-14)
    Answer: pH = 9.03 ( 9.03 )
    10.What is the pH of a solution of 0.36 M NaF? Calculate the percent hydrolysis. (Ka = 6.7 x 10-4; Kw = 1 x 10-14)
    Answer: pH = 8.37; 6.44 x 10-4 % ( 8.37; 6.44 x 10-4 % )
    11.Calculate the dissociation constant of the conjugate base of HC2H2ClO2. Ka = 1.36 x 10-3.
    Answer: 7.35 x 10-12
    12.What is the pH of an aqueous solution of 0.1 M CH3CH2NH3Cl? ( Given: Kb = 4.28 x 10-4; Kw = 1 x 10-14)
    Answer: pH = 5.82 ( 5.82 )
    13.What is the dissociation constant of the conjugate acid of NH3. Kb = 1.8 x 10-5.
    Answer: 5.56 x 10-10
    14.Find the pH of a 0.49 M NaCH3COO solution. (Ka for acetic acid is 1.8 x 10-5; Kw = 1 x 10-14)
    Answer: pH = 9.22 ( 9.22 )

    Aqueous Solution of a Weak Acid And Its Salt or a Weak Base And Its Salt

    Calculation of [H3O+] or pH of a buffer solution is accomplished using the dissociation constant equation of a weak acid or a weak base.

    For an aqueous solution containing a weak acid and its salt:

    Inverting the last term of the equation gives us: The last equation is the general form of the Henderson-Hasselbach equation.
    15.270 mL of 0.03 M NaOH solution is added to 270 mL of 0.04 M HF solution. Find the pH of the resulting solution. Given: K for hydrofluoric acid is 6.7 x 10-4.
    Answer: pH = 3.65 ( 3.68)
    16.If 3 L of 0.02 M CH3NH2 solution has a pH of 10.41, find the weight of CH3NH3Cl dissolved in the solution. ( Kb for methylamine is 4.8 x 10-4; mol. wt.methylamine = 67.5181 g/mol )
    Answer: 7.57 g
    17.What is the pH of an aqueous solution having a concentration of 0.03 M HF and 0.02 M NaF? ( Ka = 6.7 x 10-4)
    Answer: pH = 3.00 ( 3.03 )
    18.An aqueous solution of 0.11 M HNO2 and 0.11 M NaNO2 has a volume of 530 mL. If 1.2 g of NaOH is added to this solution, find the change in pH. Assume no change in the volume of solution. ( Ka = 4.5 x 10-4; NaOH = 39.99707 g )
    Answer: pH = 3.84 ( 3.84)
    19.An aqueous solution has a strength of 0.07 M HF and 0.04 M NaF. Calculate the pH of this solution. ( Ka = 6.7 x 10-4)
    Answer: pH = 2.93 ( 2.95 )
    20.In a buffer solution of 0.07 M C6H5COOH and 0.07 M NaC6H5COOH, HCl is added to it such that its calculated concentration in the solution is 0.06 M. Assuming a constant volume of solution, find the change in pH of the solution. ( Ka = 6.3 x 10-5)
    Answer: pH = 3.09 ( 3.12 )
    Posted by Gerardo Emerzona I