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Weak Base Equilibrium (continued)

Weak Base Equilibrium (continued). HA H + + A -. HCN H + + CN -. Every weak acid has a corresponding weak conjugate base. eg:. Here, HA is a weak acid, and A - must be a weak base. (it also happens to be the conjugate base of HA).

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Weak Base Equilibrium (continued)

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  1. Weak Base Equilibrium (continued) HAH+ + A- HCN H+ + CN- • Every weak acid has a corresponding weak conjugate base. eg: • Here, HA is a weak acid, and A- must be a weak base. (it also happens to be the conjugate base of HA) eg: • If HCN is a weak acid, then CN-must be a weak base. • As weak acids get weaker (going down the KA table), their corresponding conjugate bases get stronger.

  2. H3PO4H2PO4-+ H+ H2PO4-H++ HPO42- H2PO4-+H+ H3PO4 eg: • H3PO4is a weak acid, thus, H2PO4- is a weak base. But… • H2PO4- could also act as a weak acid (it still has H+ left), it is an amphiprotic substance. • A substance is amphiprotic if it has hydrogen to donate and also has the capacity to accept it. eg: acid base eg: H2CO3 HCO3- CO32- acid only amphiprotic base only

  3. [H3O+][A-] [HA][OH-] [HA] [A-] HA+ H2OH3O+ + A- A-+ H2OHA+ OH- • Relationships between KA & KB for any conjugate acid base pair: (eg: HA & A-) generic acid KA= conjugate base KB=

  4. [HA][OH-] [H3O+][A-] [A-] [HA] • Multiply KA by KB: KAx KB = = [H3O+][OH-] KAx KB= KW (where KA by KB are for a conjugate acid base pair) ie: Given KA for an acid, you can calculate KB for its conjugate base. (@ 25C) KW = 1.0 x 10-14

  5. KW 1.0x10-14 1.0x10-14 KW 6.2x10-8 KA (H2CO3) 4.3x10-7 KA (H2PO4-) eg: Calculate KB for each of the following: a) HCO3- KAx KB= KW = 2.3 x 10-8 KB= = b) HPO42- = 1.6 x 10-7 KB= =

  6. H2PO4-+ H2OH++ HPO42- H2PO4-+ H2OH3PO4+ OH- KW 1.0x10-14 KA (H3PO4) 7.5x10-3 eg: Calculate the pH of a 1.0M solution of H2PO4-. First we need to know if H2PO4- acts as an acid or a base. (KA) or (KB) Compare KA with KB KA = 6.2x10-8 = = 1.3 x 10-12 KB = Thus, although H2PO4- is an amphiprotic substance we see that KA>KB and it is more likely to act as an acid.

  7. [H3O+][HPO42-] x2 [H2PO4-] 1.0 - x Finally, we use KA= 6.2x10-8 = (assuming x is negligible) x = 2.5x10-4 [H3O+] = 2.5x10-4 pH = 3.60

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