Relationship Between Ka and Kb
Consider the dissociation of a weak acid: CH 3 COOH (aq) + H 2 O (l) H 3 O + (aq) + CH 3 COO - (aq) K a = [H 3 O + ][CH 3 COO - ](the larger the k a value, the [CH 3 COOH]stronger the acid) In water the acetate ion acts as a base: CH 3 COO - (aq) + H 2 O (l) OH - (aq) + CH 3 COOH (aq) K b = [OH - ][CH 3 COOH](the larger the k b value, the [CH 3 COO - ]stronger the base)
The product of K a and K b gives an interesting result: [H 3 O + ][CH 3 COO - ] [OH - ][CH 3 COOH] [CH 3 COOH] [CH 3 COO - ] K a K b = = = The larger the [H 3 O + ], the smaller the [OH - ] and vice versa Therefore, the conjugate of a strong acid is always a weak base and the conjugate of a strong base is always a weak acid. ie.) Large K a value (strong acid, weak conjugate base) Small K a value (weak acid, strong conjugate base) [H 3 O + ][OH - ]KwKw 1.0x K a K b =
Ex) Calculate the pH of a solution that contains 12.5g of sodium acetate (NaCH 3 COO) dissolved in 1.0L of H 2 O. Only the acetate ion affects the pH of the solution. (K a for acetic acid is 1.8x10 -5 ) NaCH 3 COO (aq) Na + (aq) + CH 3 COO - (aq) CH 3 COO - + H 2 O CH 3 COOH + OH - (aq) iceice M x + x xxx n = m/M = (12.5 g)/(82g/mol) = mol C = n/V = ( mol)/(1.0L) = M
Can we simplify? /Kb > 500YES x 2 = 5.56x (0.1524) x 2 = 8.47x x = 9.2x10 -6 [OH-] = x = 9.2x10 -6 pOH = 5.04 pH = 8.96 K b = K w = 1x = 5.56x K a 1.8x10 -5 K b = (x)(x) = 5.56x ( x)
Ex) What is the pH of a 0.35 M NH 4 Cl solution? Only the ammonium ion affects the pH of the solution. (K b for ammonia is 1.8x10 -5 ) NH 4 Cl (aq) NH 4 + (aq) + Cl - (aq) NH H 2 O NH 3 + H 3 O + iceice 0.35 M x + x xxx 0.35 M K a = K w = 1x = 5.56x K b 1.8x10 -5
Can we simplify? 0.35/Kb > 500YES x 2 = 5.56x (0.35) x 2 = 1.946x x = 1.39x10 -5 [H 3 O + ] = x = 1.39x10 -5 pH = 4.86 K a = (x)(x) = 5.56x (0.35 – x)
HOMEWORK p409 #35-38