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Understanding Acid and Base Dissociation Constants: Calculating [H+] for Weak Acids

This guide explains how to calculate the concentration of hydrogen ions, [H+], for weak acids using the acid dissociation constant (Ka). Unlike strong acids that fully dissociate, weak acids do not ionize completely, making the starting concentration of the acid not equal to [H+]. We will explore equilibrium expressions, the significance of Ka values, and utilize ICE tables to determine [H+] for weak acids. Additionally, the concept of percent dissociation will be introduced to illustrate the fraction of acid that has ionized.

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Understanding Acid and Base Dissociation Constants: Calculating [H+] for Weak Acids

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  1. Ka, Kb Acid and Base Dissociation Constants

  2. How do we calculate [H+] for a weak acid? • We know that strong acids dissociate 100% and that, therefore, the [H+] equals that [acid] that we start with • What about weak acids? • Don’t ionize 100%, so the [H+] is NOT the same as our starting concentration of our acid!

  3. Recall Kw • Autoionization of water: H2O ↔ H+ + OH- Or H2O + H2O ↔ H3O+ + OH- • Keq = • Kw = [H3O+] [OH-] = 1.0 x 10-14

  4. Ka • Similarly to Kw we can write an equilibrium expression for the dissociation of a weak acid • The equilibrium for a weak, monoprotic acid (HA) looks like this: HA(aq) + H2O(l)↔ H3O+(aq) + A-(aq) • So, we can write an equilibrium expression that looks like this: where Ka is the acid ionization constant

  5. Ka • Example: Write the equilibrium expression for the ionization of acetic acid. CH3COOH(aq) ↔ CH3COO-(aq) + H+(aq)

  6. Ka and Acid Strength • Ka values are typically between 1 – 1 x 10-16 • The higher the value of Ka, the more the acid dissociates in water and, hence, the stronger the acid

  7. What about weak bases? • Weak bases also form an equilibrium in water: B(aq) + H2O(l) ↔ HB+(aq) + OH-(aq) • This can be represented by the base dissociation constant: • Like Ka, a higher Kb means that more B has dissociated and, therefore, the stronger the base

  8. Note: Coefficients and Equilibrium Expressions • If you have coefficients in your reaction equation, they become subscripts in the equilibrium expression: 2AB → A2 + B2

  9. Try it • Write the equilibrium expression for the dissociation of NH3 in water. NH3(aq) + H2O(l) ↔ NH4(aq)+ + OH-(aq) Try the Self Test 10.2

  10. So how does Ka help us find the [H+]? • The Ka’s for almost every weak acid you could think of have been measured (at 25oC) and recorded • If we know the value of Ka and the starting concentration of our weak acid, we can solve for [H+]

  11. Try It: • What is the concentration of H+ in 0.50M HF at 25oC? • From the acid table, Ka = 7.1 x 10-4, so: HF(aq) ↔ H+(aq) + F-(aq) • Now what? Now, we use ICE tables!

  12. ICE table • HF(aq) ↔ H+(aq) + F-(aq)

  13. Solve for x • In this case x is our [H+]

  14. Short Cut: • If < 500, the change in the initial concentration (x) is negligible and can be ignored. ]

  15. Percent Dissociation (aka. Percent Ionization) • The fraction of molecules that dissociate compared to the initial concentration, expressed as a percent: • Percent dissociation = • Ex: If a 0.10 M solution of benzoic acid was found to dissociate to give a [H+] = 1.1 x 10-3 M, the percent dissociation would be:

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