An acid HX is 25% dissociated in water?

If the equilibrium concentration of HX is 0.42 M, calculate the Ka value for HX.

Please show the steps to get the answer.

1 Answer

  • 1 liter of 0.42 M HX = 0.42 mole HX

    dissociated = 25%

    in solution [H+] =[X-]= 0.42*25/100 =0.105 mole

    [HX] = 0.42 – 0.105 = 0.315 mole

    Ka = [H] [X] / [HX] = (0.105*0.105)/0.315 = 0.035

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