Relative strength of acids and bases

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 k_a/k_b expressions to construct expressions for the equilibrium constant of two competing conjugate pairs.
• How to calculate the equilibrium constant with two competing conjugate pairs.

• In Acid dissociation constant we looked at the expression for k_a (and k_b), which tells us how dissociated a weak acid (or weak base) is in solution. The larger the K_a (or K_b) value, the stronger the acid (or base).
  That the K_a value is basically an “acid 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 K_a and K_b 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 K_a values¹ will identify which it is: K_a (CH₃COOH) = 1.4 * 10^-5, while K_a (H₃PO₄) = 6.9 $*$ 10^-3. This shows that H₃PO₄ donates protons with greater ability than CH₃COOH.
    Therefore an equilibrium can be written:
    
    H₃PO₄ + CH₃COO^- $\rightleftharpoons$ CH₃COOH + H₂PO₄^-
    The K_a values show, H₃PO₄ is the stronger acid which means the equilibrium should   favor the products. We can write an equilibrium constant expression for this:
    
    K_eq = $\frac{\left[CH_3COOH\right] \left[H_2PO_{4}^{-} \right] }{ \left[H_3PO_4\right] \left[CH_3COO^{-} \right] }$
    
  • By multiplying through by [H⁺], the following expression is obtained:
    
    K_eq = $\frac{\left[CH_3COOH\right] \left[H_2PO_{4}^{-} \right] \left[H^{+} \right]}{ \left[H_3PO_4\right] \left[CH_3COO^{-} \right] \left[H^{+} \right]}$
    While:
    K_a (H₃PO₄) = $\frac{ \left[H_2PO_{4}^{-} \right] \left[H^{+} \right]}{ \left[H_3PO_4\right] }$
    And:
    K_a (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 K_a expressions!
    
    K_eq = K_a (H₃PO₄) $*$ $\frac{1}{K_a (CH_{3} COOH)}$
    
  • Simplified further:
    
    K_eq = $\frac{K_a(H_3PO_4)}{K_a (CH_{3} COOH)}$
    
  • 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:
    
    K_eq = $\large \frac{6.9 \, * \, 10^{-3}}{1.4 \, * \, 10^{-5}}$ = 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 K_a value for both conjugate acids then set up the equilibrium and K_eq expression like this:
    
    Where:
    HX = conjugate acid (stronger acid; larger K_a),
    HY = conjugate acid (weaker acid; smaller K_a),
    X^- = conjugate base of HX
    Y^- = conjugate base of HY
    
    HX + Y^- $\rightleftharpoons$ HY + X^-
    K_eq = $\large\frac{[Proucts]}{[Reactamts]}$ or:
    K_eq = $\large\frac{K_a (HX,\; acid\; in\; reactants)}{K_a (HY,\; acid\; in\; products)}$
    
  • If done correctly, the K_eq 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.