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  3. consider the diprotic acid h2x, where ka1 = 1.75 x

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 Ka1= 1.75 x 10-5 and Ka2 = 4.3 x10-12 at 25 °C. The concentration of HX- (aq)in a 0.125 M solution of H2X at 25 °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
        H2X ----------------------------> HX- + H+

I       0.125                                     0          0

C       -x    +x +x

E   (0.125-x)                                   x          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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