Question

13. Tartaric acid, H:C:H:04, is a diprotic acid with Kal = 9.20*10* and K2= 4.31x10. Determine the pH and the concentrations of all three forms of tartaric acid in a 0.261 M solution.
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Answer 1

Dipostic acid (H2 C4 HAOA) : 0.261 m и, са и, од исцидо, Сва) + н* ta) 0-260 Ба, д.2x10"4 н от - Х + x О. 26) - ка. - См"] [ис, и, од ] . (и, ди, 4) x б261 - x ka, is very small, so, 0.261-2 a 0.261 х Кар - О. 261 2) 9.2 х 04 - х* О 26 х? - 2 -4012 x to 4 х = 1,5 49 х 10- - 2 м - Си4 04 И4) - 1.49 х 10 си"] = 1349 10 м, Сис, 0, 1, 2 - 149 хтог*м }, {49 х 10 м

at equilibrium, [12 CQ HA Oq] = 0.261-1 = 0.261 - 1.549 x 10-2 = 0.261 -0.01549 = 0.24551 . [4₂ KG Hq Oq J = 0.24551 M

pH = -log Cut] = -log (1.549 * 182) = 1.80 IPH = 180 Now other reaction. + 47 ka = 4-31x 10-s HEAHOO I 1.549810-2 c -k E 1.549x10²_x 1.549x102 the ta 1:5498102 + x [ut] [ Caha0aJ [HC Haag X (1.549 x10²+2) (1.549810²-2) Here ka, is very small, so, neglect a because of less dissociation. Hence, 1.5498102 » = x ka = x (1.549*102) (1-549 XT0-2) x = 4:31x10-5 ..[CAHAOR] = 4.31x105 m [ucana Oâ ] = 1.549x102x = 1.549x102 m