1 / 7

Understanding Weak Acids, Conjugate Bases, and Amphiprotic Substances in Equilibrium

This document delves into the relationships between weak acids and their conjugate bases, illustrating how the strengths of these species vary with acid strength. It elucidates the concept of amphiprotic substances like H2PO4-, which can act as both an acid and a base. The relationships between the acid dissociation constant (KA) and the base dissociation constant (KB) are explored, along with practical calculations for finding KB from given KA values. The document also describes how to determine the pH of solutions involving amphiprotic species.

zia-chang
Télécharger la présentation

Understanding Weak Acids, Conjugate Bases, and Amphiprotic Substances in Equilibrium

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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

More Related