Question

Determine the pH of each aqueous solution

. a. 0.475 M oxalic acid (C2O4H2) and 0.500 M HClO.

b. 0.150 M sodium formate (NaHCO2) and 0.138 M pyridine (C5H5N)

Answer 1

o 0.475 Coon oxalic acid ka, = 5.9x102 kq2 = 6.48105 Соо и acid is 0.500M HUO K9 3.0X10-8 oxalic acid Hoo both are weak acid but oxalic Shonger than HUO since oxalic acid is diprotic acid but ka, kaa the H* bom first dissociation constant only. so consider Cook +420 → COO + H+ COOH initial 0.475 (M) Change - - At 0.475-02 epuilibrium ka, = [cool] [ut] [(2004), 5.9 X10-2 = 0.475-x 2.803x10^2 - 5-9410-2x = 12 X +5.9x102x -.803x10^2 = 0 compare this equation with ax²+bx+c=0 a = 1 b = 5.9x10-2 (= -2.803x16-2

= -bt b²-490 22 = -5.9x10^2 + 1 (5-4 X10-43 + 4x 2.803x102x1 -5.9x10* £ 34.81x10-4+11.212x102 -5.9 X10 1.115601 = 0.1405 X = 0.1405 x= [4+] = 0.1405M belog NOW 1 H2O + Huo initial 0.5 ² H+ 0.1405 + co- o Change A+ 0.5-2 0.1405th a equilibrium Ka= [ht] [40] [no] 3.0 X10-8 = x (x +0.1405) a is very small Co.s-x) ng leta 3.0x108 - x) x0.1405 - 0.5 1.07X107 = x so CM+] = 0.1405+ 1.07x107- 0.1405M bu= -log [H+] = -log (0.1405) = 0.85 Aruwer

bi 0.150M Sodium formate (NaHCO3) and 0.138 M Pyridine kg of HCO3 = 2.088104 Kb J G H N = last 1.5X10-9 H605 is strong base dis reaction as follow as compare to Pyridine so 2 HCO 3 + oH inisial Change Ho5 theo 0.150 -x At equieli brium 0.150-2 kb= [12203] [ok] Laos ) - 030) Coloz) ko = (1) (a) (0.150->) 2.08X104. r? X xloo L51 0.150 neglecta in denominator 0.150- 0.312 x10-4 = 22 0.0056 = x 0.0056 x100 = 3.7 5518 0.150 so Approximation railed COH] = 2 = 0.0056 M

bi 0.150M Sodium formate (NaHCO3) and 0.138 M Pyridine kg of HCO3 = 2.088104 Kb J G H N = last 1.5X10-9 H605 is strong base dis reaction as follow as compare to Pyridine so 2 HCO 3 + oH inisial Change Ho5 theo 0.150 -x At equieli brium 0.150-2 kb= [12203] [ok] Laos ) - 030) Coloz) ko = (1) (a) (0.150->) 2.08X104. r? X xloo L51 0.150 neglecta in denominator 0.150- 0.312 x10-4 = 22 0.0056 = x 0.0056 x100 = 3.7 5518 0.150 so Approximation railed COH] = 2 = 0.0056 M

consider for Pyrindine ² C5 H5 Nht - initial change CGNHS + H₂O 0.138 - tot 0.0056 At 0.0056+2 equilibrium 0.138-X ko= [cghsntu] [ot] [Chor] 1.5x109 = (x) (0.0056th] neglecta [0.138-2) 1.5 X10-9 = (2) (0.0056) 0.138 3.7x10-8 x [OH-] = 0.0056+ 3-7x108 = 0.0056 bok= -log (OH) s-log (0.0056) bon = 2.25 þht bon = 14 bn= 14-2.25 = 11.75 14=11.75 Answer