Question

The K, of a monoprotic weak acid is 0.00481. What is the percent ionization of a 0.154 M solution of this acid? percent ionization:

Answer 1

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Answeyi ginen, ta = 0.00481 = 4.818 10-3 Let Ha be the mono protic weak acid. - HA + H₂O H₂ot + A- Initial 00154 eqb 0.154-1 ka - Гној ГА] CMAJ let sonde og ha dissotiate at equilibrium * ka = non 0.154-a 212 + 4.81 X10-3 - 0.154-1 = 32 + 0.854 X 4.81x1013 - 4.81x10 32 - 0 -0.64372X10 2 + 4.81 % 103 a = -0.0297 12 = 0.0249 concentration can't be negatine, hence My = -0.0297 es neglected. .7272 = 0.0 249. [H₂Ot] = 0.0 249.

%. Ionization = [H3O+] x 100 [HA] = 0.0249 x 100 0.154 = 0.16 16 x 100 = 16.16 Y

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