A monoprotic weak acid, when dissolved in water, is 0.92% dissociated and produces a solution with pH = 3.42. Calculate K_a for the acid. A) 1.4 × 10^-7 B) 2.8 × 10^-3 C) 3.5 × 10^-6 D) We need to know the initial concentration of the acid. E) none of these

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We can use the following equation to solve for the dissociation constant of the weak acid: Ka = [H+][A-]/[HA] We know that the concentration of the acid at equilibrium ([HA]) is equal to the initial concentration minus the concentration that has dissociated: [HA] = [HA]0 - [HA]diss We also know that the percent dissociation is equal to the concentration that has dissociated divided by the initial concentration, multiplied by 100: % dissociation = ([HA]0 - [HA])/[HA]0 x 100 Putting this all together, we can set up the following quadratic equation to solve for Ka: Ka = x^2/([HA]0 - x) Where x is the concentration of H+ and A- at equilibrium, and [HA]diss is also equal to x. Since we know the pH of the solution (pH = 3.42), we can solve for the concentration of H+ using the equation: pH = -log[H+] 3.42 = -log[H+] [H+] = 4.4 x 10^-4 M We also know from the percent dissociation that: 0.92% = (x/[HA]0) x 100 0.92/100 = x/[HA]0 x = 0.0092[HA]0 Substituting these values into the quadratic equation for Ka: Ka = (4.4 x 10^-4)^2/[HA]0-0.0092[HA]0 Ka = 1.87 x 10^-9 Finally, since we are looking for K_a, we can use the relationship: K_a = Kw/Ka Plugging in the value for Ka that we just solved for: K_a = (1.0 x 10^-14)/(1.87 x 10^-9) K_a = 5.35 x 10^-6 The closest answer choice is C: 3.5 x 10^-6, so that is our answer.

A Monoprotic Weak Acid, When Dissolved in Water, Is 0