The first and second ionization constants ofH 2 CO 3 are 4.2×10 -7 and4.8×10 -11
ID: 687969 • Letter: T
Question
The first and second ionization constants ofH2CO3 are 4.2×10-7 and4.8×10-11. Calculate the concentrations ofH+ ion, HCO3-,CO32- and undissociatedH2CO3 in a 0.080 MH2CO3 solution Matching pairs: [H2CO3] — [HCO3-] — [CO32-] — [H+] — the answer choices for each have these macthing pairs with ofthese could be the answer 0.080M 1.83 * 10 ^-4M 4.8*10^-11M Matching pairs: [H2CO3] — [HCO3-] — [CO32-] — [H+] — the answer choices for each have these macthing pairs with ofthese could be the answer 0.080M 1.83 * 10 ^-4M 4.8*10^-11M Matching pairs: [H2CO3] — [HCO3-] — [CO32-] — [H+] —Explanation / Answer
H2CO3 is a WeakAcid So it cannot decompose completely . The concentration of the given species arecalculated by using ICE table H2CO3 (aq)<---------> HCO3- (aq)+H+(aq) I(M) 0.08 0.00 0.00 C(M) -x +x +x E(M) 0.08-x x x Ka1 =0.08-x 4.2×10-7 =x2/0.08-x x =1.83*10-4 M So the concen tration ofthe H+ ion, HCO3- matcheswith x =1.83*10-4 M Onceagain HCO3- (aq) is alsodecompose in the same way. HCO3-(aq)<---------->H+(aq)+CO32- (aq) I(M) 1.83*10-4M 0.00 0.00 C(M) -x +x +x E(M) (1.83*10-4 -x) x x Ka2 =x2/ (1.83*10-4 -x) (4.8×10-11) =x*1.83*10-4 -x) / (1.83*10-4 -x) Therefore[CO32- ] =4.8×10-11 The concentration of the given species are [H2CO3] = 0.080M [ HCO3-]=1.83*10-4 M [CO32-] = 4.8×10-11 [H+] =1.83*10-4 M The concentration of the given species are [H2CO3] = 0.080M [ HCO3-]=1.83*10-4 M [CO32-] = 4.8×10-11 [H+] =1.83*10-4 MRelated Questions
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