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

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

Expert Solution

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

queen_honey_blossom answered 1 year ago

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