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10.6: The pH of Weak Acid Solutions

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    79596
  • For a solution of a strong acid, calculating the [H3O+] concentration is simple; because the acid is 100% dissociated, the concentration of hydronium ions is equal to the molar concentration of the strong acid (this is, of course, only true for a monoprotic acid such as HCl or HNO3; for H2SO4, [H3O+] = 2 × [H2SO4], etc.). For a weak acid, however, the hydronium ion concentration will be much, much less than the molar concentration of the acid and [H3O+] must be calculated using the value of Ka. We can approach this using an ICE table, like we did for previous equilibrium problems. If we prepared a solution of acetic acid that was exactly 0.50 M, then initially [CH3COOH] is 0.50 M and both [CH3COO] and [H3O+] are zero. A small amount of CH3COOH will ionize; let’s call this x, making the change for [CH3COOH] “-x”, increasing both [CH3COO] and [H3O+] by the amount “+x”. Finally, the equilibrium concentration of [CH3COOH] will be (0.50 M – x) and both [CH3COO] and [H3O+] will be x. The completed table is shown below.

     
    [CH3COOH]
    [CH3COO]
    [H3O]+
    === Initial ===
    0.50 M
    0
    0
    Change
    - x
    + x
    + x
    Equilibrium
    0.50 M - x
    x
    x

    The expression for Ka for acetic acid is given in equation in section 10.5. Substituting for our equilibrium values:

    \[K_{a}=1.8\times 10^{-5}=\frac{[H_{3}O^{+}][CH_{3}COO^{-}]}{[CH_{3}COOH]}=\frac{x^{2}}{0.50-x}\]

    \[x^{2}+9.0\times 10^{-6}x-1.8\times 10^{-5}=0\]

    The above equation is a quadratic equation and we could solve it using the standard quadratic formula. This is not necessary, however, because acetic acid is a weak acid and by definition, very little of the dissociated form will exist in solution, making the quantity x very, very small. If x is much, much less than 0.50 M (our initial concentration of acetic acid), then (0.50 M – x)  0.50 M and the equation simplifies to:

    \[K_{a}=1.8\times 10^{-5}=\frac{[H_{3}O^{+}][CH_{3}COO^{-}]}{[CH_{3}COOH]}=\frac{x^{2}}{0.50}\]

    \[x=[H_{3}O^{+}]=\sqrt{(1.8\times 10^{-5})\times 0.50}=3.0\times 10^{-3}M\]

    We can test our assumption by substituting for x; (0.50 – 0.0030) = 0.497, which rounds to 0.50 to two significant figures. Because the concentration of hydronium ion is very small for a weak acid, for most typical solutions, the concentration of hydronium ion can be estimated simply as:

    \[[H_{3}O^{+}]=\sqrt{(K_{a}\times C_{0}}\]

    where C0 is the initial molar concentration of the weak acid.

    Exercise \(\PageIndex{1}\)

    1. Nitrous acid (HNO2) is a weak acid with a Ka of 4.3 × 10-4. Estimate the hydronium ion concentration and the pH for a 0.50 M solution of nitrous acid in distilled water.
    2. Acetic acid is a weak acid with Ka = 1.8 × 10-5. For a solution of acetic acid in water, the [H3O+] is found to be 4.2 × 10-3 M. What is the concentration of unionized acetic acid in this solution?

    CH3COOH(aq) + H2O(l) ⇄ CH3COO(aq) + H3O+(aq)

    1. A solution is prepared in which acetic acid is 0.700 M and its conjugate base, acetate anion is 0.600 M. As shown above, the Ka of acetic acid is 1.8 x 10-5; what will the pH of this solution be?
    2. What concentration of the weak acid, acetic acid (Ka = 1.8 × 10-5) must you have in pure water in order for the final pH to be 2.38?

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