Question

What is the molar concentration of carbonate ion in a 0.10 M solution of carbonic acid?

  Ka1 = 4.3 × 10–7 ; Ka2 = 5.6 × 10–11

Please report the answer in units of M (do not write unit, only number)

Write numbers as 1.23E-5

Answer 1

H2CO3 +420 ² HCOğ + H₂ot ka, = 4.3x10 7 HCO3 + H2O È což +4got Rag= 5.6x10-" consider first dissoci ation g 4,603 H₂CO3 + H2O ² HCO3 +H307 initial 0.1 Change 2 -X 0.1- At epuilibrium ка ka= [+603] [43ot] у, соз) 4.3x1017 = (426033 )(2) 01-X i$ * X100 <51 neglectxe in denominator 4.3410-7 = x² 0.000207x10. 5-1. OL Bo Approximation valid 0.1 0.43x10 7 = n² 0.000 207 = x so [4*) - [4 C03 ]=% = 0.000207 Now second dissociation HCO3 +420 ² co3² inisial 0.000207 +4+ 0.000207 -X m+0.000207 chя ти 0.000207-2 At Guilibrium

S Ka= [co3²] [430f) (मी 5.6X10!! = 1 0.000207 +) (v) 10.000207-2) x is small so neglect 5.6x10-11 = - 0.000204) x 0.000 got x= 5.6810-11 [ 1032-] = 5.6810-11 ser