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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.
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Anya Stehr

Feb 20, 2025

1 liter of 0.42 M HX = 0.42 mole HXdissociated = 25%in solution [H+] =[X-]= 0.42*25/100 =0.105 mole[HX] = 0.42 – 0.105 = 0.315 moleKa = [H] [X] / [HX] = (0.105*0.105)/0.315 = 0.035... Show More
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