A2 AQA CHEM4 3.4.3

3.4.3 Acids and Bases

Brǿnsted-Lowry acid-base equilibria in aqueous solution

·         Acidsà proton donors

·         Basesà proton acceptors

·         Acid-base equilibria involve the transfer of protons (i.e. something is acting as the acid and something is acting as the base)

·         E.g. HA  + H₂O à H₃O⁺ + A^-

o   The water is acting as the B-L base

o   The HA is acting as the B-L acid

o   The H₃O⁺ is the conjugate acid

o   The A- is the conjugate base

·         Water can act as an acid and a base so it is amphoteric

·        Remember H₃O⁺ is what we call an oxonium ion

Definition and determination of pH

·         pH= -log[H⁺]

·         [H⁺]= 10^-pH

·         When an acid is diluted

o   New [H⁺] = original [H⁺] x (volume of acid/ total volume of solution)

·         A strong acid dissociates completely in solution

o   E.g. HCl à H⁺ + Cl^-

·         Monoprotic acid= donates one proton  

o   [acid] = [H⁺]

·         Diprotic acid= donates two protons

o   [acid] x 2 = [H⁺]

·         Strong bases dissociate completely in solution

o   E.g. NaOH à Na⁺ + OH^-

o   [base] = [OH^-]

The ionic product of water K_w

·         Water is weakly dissociated

o   i.e. H₂O <--> H⁺ + OH^-

·         At a constant temperature [H⁺]=[OH^-] and this means water is always neutral at a constant temperature

·         K_w= [H⁺][OH^-]

·         At 25^oC, K_w is 1 x 10^-14 mol²dm^-6

·         We can rearrange K_w to get [H⁺] as the subject and sub in our [OH^-] to find out the [H⁺] and use this to find the pH of a base

·         Since [OH^-]=[H⁺] we know that K_w= [H⁺]² so √K_w = [H⁺] and again we can use this to find pH

·         The dissociation of water is endothermic so the value of K_w increases and temperature increases and since the position of the equilibrium shifts right and the conc of H⁺ increases, the…