Question

pH of Weak Acids 19. Consider the following acids; CH5COOH ( benzoic acid) Ka = 6.14x10-5 H3BO3 (boric acid) Ka = 5.83x10-10 H2P04 (dihydrogen phosphate) Ka = 6.30x10-8 HNO3 (nitric acid) >>> 1 Put the acids in order from strongest to weakest. 20. Calculate the pH and percent ionization of the following solutions (Use ICE for weak acid calculation) a. 0.15 M HNO3 0.82; 100% b. 0.15 M HNO2 ( ka = 4.5x10-5) 2.59; 1.71%

Answer 1

19. Acids and their ka values are - CH3 COOH (benzoic acid) Ka = 6. 14x10 5 H₂B03 (boric acid) ka = 5.88 x 10-10 H₂PO4 (dihydrogen phosphate) ka = 6.30 x 108 HNO3 (nitric acid) kan Ka value indicates the dissociation const of acid. Higher the ka value, higher in the dissociation of the acid i.e. formation of Htion is more, so higher in the acidity of that acid. The order of ka value of the above acids is a Ka (HNO3) ? Ka (CC Hs 200H)) Ka (H2PO2) I ka (H₂ BO3) So, the acidity order will be same as that of the ka order. Therefore, the order from strongest to weakest acid is - HNO3) C6H5COOHY H₂PO4.7H₂ BOB 20. a) conc. of HNO₃ = 0.15 M. HNO3 is strong acid, so it undergoes complete dissociation. - HNO - Ht+ NO 0.15(M) 0:15(M) 0.15(M) . [HA ] = 0.15 (M) we know, for strong acid, pH = -log [ht] an log (0.15) = 0.82

As, it underges complete dissociation, so degree of dissociation (&) = 1x100 % - 100% che sto ionization) b) conc. of HNO = 0.15M. Ka of this acid – 4.5x10-5 The ka value indicate that it is not strong acid, so it undergoes partial dissociation. for such type of acid - pH = pka - ғас = √ { log ka} - I loge I 57-log (4.5X1095)? - log 0.15 = 2.59 ret, degree of dissociation in a for HNO HNO, H+ HNO C (1-0) ac ac a conc. of UNO ka = [ht] [NO2 [4NO2J - acxac (1-9) . ~ xrc [.& is be very deas, so 1-a ni - 4.58105 0.15 -0.0123 21.73% 2. Percent ionization of HNO is 1.73%