Question

Calculate the pH and the equilibrium concentrations of HC2O4- and C2O42- in a 0.1160 M oxalic acid solution, H2C2O4 (aq). For H2C2O4, Ka1 = 5.9×10-2 and Ka2 = 6.4×10-5

pH =

[HC2O4-] = M

[C2O42-] = M

Answer 1

Soli coolt Given: {coch = col16 m., Kapas.9x10? oxalic acid ka = 6.4x 10 Find play = ?? (Now) Ist -2 (ka, a Seax 10 cre59) ionization ~ çoch coot Can cooth (ad) Initial) o.llom Equin) 0.116- - ka = CH] [cool] 2 COCH - 0.0592 rooms 0-116-4 >> »?toorax - 02006844 = 0 solving Quadratic equation using x - 7 for ax²+bx+c=0 gives (x = 0.0583311 za coo ~ (Hoy ] = x= 0.0593m Carta a aa 0.0 583m at kai
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and ionization: (ka,- 6.4 x 10 coolt cap Coo Casa 0.05830 Initial) o 0.0583 0-0583+y Equi 0.0883 - y a ko -) -8 y0.0583+ y) 6.4 X 10 = (0.0583-y) Coch Y - - - y = 6.4x10 since kan small 0.0583+ y 2 ~0.0583 & - 3 6-48103 - Yo0585) 20883) (small) (or) (20) = 4= 6.4x105m & Total Cett} = x+y ~x~ 0.0583 .PH – log[+] = – lag (0.0 582) 21-23 PH 1 = 1.23 O 24 0583M 4 = 6.4810 m.
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