Question

Consider the diprotic acid H₂X, where K_a1 = 1.75 x 10^-5 and K_a2 = 4.3 x 10^-12 at 25 °C. The concentration of HX^- (aq) in a 0.125 M solution of H₂X at 25 °C is approximately

• 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 1

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