Question

For a 7.95x10-2 M solution of H,C0g, calculate both the pH and the CO2-ion concentration. State symbols are not shown for water or the aqueous species in the reaction equations. HỌCO, + HọO 2 Hyo* + HCO, HCO3 + H20 H30* + CO32- Ka = 4.2x10? K22 = 4.8*10-11 pH = [C0321 = M

Answer 1

Ans: The The reaction stated is as follows... reaction stated este H₂ CO₂ + H₂O H₂O + + HCO₃ ska, =.4.2010 HCO3 + H₂O H₂O + + co ; ka, 4.8*10-11 Now; the equilibrium constants can be explicitly expressed as; Ka, = [H] [ HCO3]; where [H +] = [Hot] [H₂ CO 3 - and , ka = [H + 1 [CO3² - [H Coz -] 11000, taking produce of two eqm ebutants, Kaj, ka, - [H+],[H005)[ht][C032] [H₂ col [Hoz] Ka, ka = [H +7². [CO3²-7 [H₂ CO]

- Teoza-] = kai. Kag. [H₂CO₃ ] 0 CH+J, ZH+] we need to Fon calculating the the pH of the soln acid. Hy CO₂ is a weak that consider Ferom ®: (H+], - ka,. [H2CO3] TH Coz >> (H+). [H203-= 4.2x10-* x 7.95X10-CM) cgiven) -> [H +),= [H 203-] = V4.2 X10- + x 7.95 X10-2 (M) = 1.82x10-4CM) 2 [...H₂CO3 + H2O5 inlil reatio] . Fecom ③ [HT] = kag . I HCO₃-] . , in this step [co ] Ht comes from dissociation of HCO3 * [HH], [2032-] = kaz. I HAD ] » [H+] = [ 203 2-7 = 4.8x10-11x 1.82 x 10-4 CM) Tresing Ear o » [ht] = [2032-= 9.36 X10-8(M) - [... HCO3- H+ + co2 in 1il reatio] This is pH - (log [44])+(-log[114] 2) Thus, pH = -log(CH+], + CH+J₂) anlog (1.82*10-4) - 3.74. Also; [CO3 2 -1 = 9.36x10 - 8 (M) [from Ear ] (similar recruit would be obtained if we use Egr ① to calculate the same).