Question

An unknown weak acid with a concentration of 0.078 M has a pH of 1.80. What is the Ka of the weak acid? O of 1 point earned 2 attempts remaining An unknown weak base with a concentration of 0.170 M has a pH of 9.96. What is the Kb of this base? O of 1 point earned 2 attempts remaining

Answer 1

To find the K of the weak acid Given that weak acid with a concentration of 0.078M solution pH = 1.80 It will dissociates HA+H,0=H,0* +A Equilibrium expression ✓ _[-(,06] [HA] First find the [H3O+] pH = -log[H,0*] [H,0*]=10-** [H20*] = 10-20 = 1.58x10-M From the dissociation equation, we know there is a 1:1 molar ratio between [H,0*] and [A-] Therefore: [A-]=1.58x10-²M [A- [40] K, - HA] (1.58x10-)(1.58x10²) K = 0.078-1.58x10-9 = 4.0x103 Hence the K of the weak acid=4.0x10

To find the K, of a weak base Given that weak base with a concentration of 0.170M solution pH = 9.96 It will dissociates B+H,O=BH+ + OH v Rot (BH* TOH-] [B] First find the pOH pH+pOH = 14 POH = 14-pH = 14 -9.96 = 4.04 To find the [OH-] POH = -log[OH-] [OH-] = 10-407 = 10-.04 = 9.12x10-M From the dissociation equation, we know there is a 1:1 molar ratio between (OH) and (BH*] Therefore(BH* )= 9.12x10-$M K. - [BH* [OH-] Кь (в) (9.12x10-)(9.12x10- ) 0.170-9.12x10 = 4.90x10 Hence the Ky of a weak base = 4.90x10