Consider the diprotic acid H2X, where Ka1 = 1.75 x 10-5 and Ka2 = 4.3 x 10-12...

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Chemistry

Consider the diprotic acid H₂X, where K_a1= 1.75 x 10^-5 and K_a2 = 4.3 x10^-12 at 25 Â°C. The concentration of HX^- (aq)in a 0.125 M solution of H₂X at 25 Â°C isapproximately
A. 2.0 x 10^-5 M
B. 4.5 x 10^-3 M
C. 1.5 x 10^-3 M
D. 3.5 x 10^-5 M
E. none of these

Answer & Explanation Solved by verified expert

4.3 Ratings (549 Votes)

first construct the ICE table and then substitute the values in the K equation and then calculate the equilibrium values as follows
Â Â Â Â Â Â Â H2X ----------------------------> HX- + H+

IÂ Â Â Â Â Â 0.125Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0Â Â Â Â Â Â Â Â Â 0

CÂ Â Â Â Â Â -x Â Â +x +x

EÂ Â (0.125-x)Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â xÂ Â Â Â Â Â Â Â Â x

Ka1 = [HX-] [H+] / [H2X]

=>1.75*10^-5 = x² / (0.125-x)

=> 1.75*10^-5 * (0.125-x) = x²

=> x = 0.001479

=> 0.0015 => 1.5*10^-3 M = [HX-] = [H+]

answer => c) 1.5*10^-3 M

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