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A 0.50 m solution of an acid ha has ph = 2.24. what is the value of ka for the acid?

Cynthia Baker

in Chemistry

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Deborah Edwards on August 11, 2018

A 0.50 M solution of a monoprotic acid HA, with a pH of 2.24, would be, in the first place, a weak acid, since it does not dissociate completely. This leaves us with an equilibrium expression: HA (aq) ⇌ H+ (aq) + (aq)Where a - is the conjugate base of the weak acid.In a study of balance, reminds us that the value of ka is the dissociation constant of the acid, and has the equation:Ka = (concentration of H+)(concentration of conjugate base)/concentration of acidWe know the concentration of H+ and A - are 10^-2.24 for the definition of a pH to be-log(concentration of H+). The concentration of the acid has gone down a bit, as it has done partially dissociates into H+ and A-, so we need to subtract 10^-2.24 0.50 for the acid concentration to account for the dissociation. The last equation would become:[H+]*[A-]/[HA] = Ka(10^-2.24) * (10^-2.24) / (0.50 - 10^-2.24) = Ka(3.31 * 10^-5) / (0.494) = KaKa = 6.70 * 10^-5


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