Question

QUESTION and Ka2-5.1 x 10. What is the concentration of C04 in a 0.37M oxalic acid, HoC)0, solution? For oxalic acid, Ka1 5.6 x 10 a. 4.3185361704295229 x 10- M b.5.1x 10 M 43439613257946945 x 103 M Od1.1864242223858685 x 10-1 M e 1.4394443372357264 x 101 M QUESTION 2

need help with question 1 please

Answer 1

For of oxalic acid Ka1 = 5.6 x 10-2 Ka2 = 5.1 x 10-5 The dissociation of diprotic acid is given by H20204 - 0.370 H+ 0 + HC,04 0 Initial Change Equilibrium 0.370 - X - = 5.6x10-2 [H*][HC204) X.X a [H,C,04] 0.370- x = x² +0.056x -0.02072 = 0 By solving, x= 0.118642 [H*]=0.118642M [HC204]=0.118642M

HC,04 will further dissociates into H+ and C,0,2 Let y be the change in the conc. of all species at equilibrium. ka2 HC204 7 H+ + C204-2 Initial 0.118642 0.118642 Change - y + y +y Equilibrium 0.118642 – y 0.118642+ y y K. =- [H*][0,0,?] (0.118642+ y).y51410-s ? [HC,0.-] 0.118642 – y Since y is very small,0.118642 - y0.118642 and neglet y? 0.118642y = 5.1x10- 0.118642 :: y= 5.1x10-5 [H*]=0.12107+ y = 0.12107+(5.1x10-) = 0.12107 [0,0,?]= y = 5.1x10--M Option C is the right answer.