Relative strength of acids and bases

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Intros
Lessons
  1. Which is the stronger acid / base?
  2. Competing conjugate acids/bases.
  3. Using Ka/Kb expressions to find Keq.
  4. Using Ka/Kb expressions to find Keq (continued).
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Examples
Lessons
  1. Find the equilibrium constant for two competing weak acids and their conjugate pairs.
    Ethanoic acid, CH3COOH and carbonic acid, H2CO3 are both weak acids. Their Ka acidity constants1 are below:
    Ka (CH3COOH) = 1.4*10-5
    Ka (H2CO3) = 4.5*10-7
    1. Identify their conjugate bases and write the equilibrium equation for the reaction of the two conjugate pairs.
    2. Use the Ka values above to calculate which side of the equation is favoured. Which is the stronger acid?
  2. Use the Ka expression to compare the strengths of two weak acids.
    Formic acid, HCOOH, and butanoic acid, CH3CH2CH2COOH, are both weak acids. Their Ka acidity constants1 are below:
    Ka (HCOOH): 1.8*10-4
    Ka (CH3CH2CH2COOH): 1.5*10-5

    Write a Keq expression using the two weak acids and their conjugate bases to explain which is the stronger acid.



    1 Source for acidity constant (Ka) data: ATKINS, P. W., & DE PAULA, J. (2006). Atkins' Physical chemistry. Oxford, Oxford University Press.
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    Topic Notes
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    In this lesson, we will learn:

    • How to determine the stronger of two acids/bases in reactions between two competing conjugate pairs.
    • How to use ka/kb expressions to construct expressions for the equilibrium constant of two competing conjugate pairs.
    • How to calculate the equilibrium constant with two competing conjugate pairs.

    Notes:

    • In Acid dissociation constant we looked at the expression for ka (and kb), which tells us how dissociated a weak acid (or weak base) is in solution. The larger the Ka (or Kb) value, the stronger the acid (or base).
      That the Ka value is basically an β€œacid strength rating” is good to remember for solutions with multiple conjugate pairs. When two conjugate pairs are mixed together in equilibrium, there are two competing acids (and bases) both trying to be an acid – trying to donate H+ (and two competing bases both trying to accept H+). You can use the values for Ka and Kb to see which one is the stronger acid/base, and which side is the equilibrium favors.
      • For example, if solutions containing CH3COOH and H3PO4 were combined, there would be two acids both competing to donate protons to other species.
      • The stronger acid will do this to a greater extent. The Ka values1 will identify which it is: Ka (CH3COOH) = 1.4 * 10-5, while Ka (H3PO4) = 6.9 βˆ—* 10-3. This shows that H3PO4 donates protons with greater ability than CH3COOH.
        Therefore an equilibrium can be written:

        H3PO4 + CH3COO- β‡Œ\rightleftharpoons CH3COOH + H2PO4-

        The Ka values show, H3PO4 is the stronger acid which means the equilibrium should   favor the products. We can write an equilibrium constant expression for this:

        Keq = [CH3COOH][H2PO4βˆ’][H3PO4][CH3COOβˆ’]\frac{\left[CH_3COOH\right] \left[H_2PO_{4}^{-} \right] }{ \left[H_3PO_4\right] \left[CH_3COO^{-} \right] }

      • By multiplying through by [H+], the following expression is obtained:

        Keq = [CH3COOH][H2PO4βˆ’][H+][H3PO4][CH3COOβˆ’][H+]\frac{\left[CH_3COOH\right] \left[H_2PO_{4}^{-} \right] \left[H^{+} \right]}{ \left[H_3PO_4\right] \left[CH_3COO^{-} \right] \left[H^{+} \right]}

        While:

        Ka (H3PO4) = [H2PO4βˆ’][H+][H3PO4]\frac{ \left[H_2PO_{4}^{-} \right] \left[H^{+} \right]}{ \left[H_3PO_4\right] }

        And:

        Ka (CH3COOH) = [CH3COOβˆ’][H+][CH3COOH]\frac{ \left[CH_3COO^{-} \right] \left[H^{+} \right]}{ \left[CH_{3}COOH\right] }

      • Notice these are now expressions of dissociation? This can now be simplified using Ka expressions!

        Keq = Ka (H3PO4) βˆ—* 1Ka(CH3COOH)\frac{1}{K_a (CH_{3} COOH)}

      • Simplified further:

        Keq = Ka(H3PO4)Ka(CH3COOH)\frac{K_a(H_3PO_4)}{K_a (CH_{3} COOH)}

      • Both of these are known constants (we used them earlier!) so we can calculate an equilibrium constant to determine which side of the equilibrium is favored; it should back up our prediction that H3PO4 dissociates more:

        Keq = 6.9β€‰βˆ—β€‰10βˆ’31.4β€‰βˆ—β€‰10βˆ’5\large \frac{6.9 \, * \, 10^{-3}}{1.4 \, * \, 10^{-5}} = 492.86...

      • This value (a ratio of around 493:1) shows the equilibrium heavily favors the products as predicted by the acid dissociation constants. For any two conjugate pairs in competition, look up the Ka value for both conjugate acids then set up the equilibrium and Keq expression like this:

        Where:
        HX = conjugate acid (stronger acid; larger Ka),
        HY = conjugate acid (weaker acid; smaller Ka),
        X- = conjugate base of HX
        Y- = conjugate base of HY

        HX + Y- β‡Œ\rightleftharpoons HY + X-

        Keq = [Proucts][Reactamts]\large\frac{[Proucts]}{[Reactamts]} or:

        Keq = Ka(HX,β€…β€Šacidβ€…β€Šinβ€…β€Šreactants)Ka(HY,β€…β€Šacidβ€…β€Šinβ€…β€Šproducts)\large\frac{K_a (HX,\; acid\; in\; reactants)}{K_a (HY,\; acid\; in\; products)}

      • If done correctly, the Keq expression will yield a value greater than 1 (showing the equilibrium shifted right; that the stronger acid dissociates more). Another way to read this is that the equilibrium will favor the side with the weaker acid.