If enough of a monoprotic acid is dissolved in water to produce a 0.0158 M solut

Question

If enough of a monoprotic acid is dissolved in water to produce a 0.0158 M solution with a pH of 6.42, what is the equilibrium constant, Ka, for the acid? You have not accounted for the autoprotolysis of water when calculating the Ka value. For weak acids that are so dilute or so weak that the pH of the solution lies between 6 and 7, the autoprotolysis of water must be taken into account when determining the Ka value. Start by writing four equations: the Kw expression the Ka expression the charge balance of the solution the material (mass) balance of the solution Use these four equations to develop an expression for Ka in terms of Kw, [H3O^+]. and the initial concentration of the acid, [HA]_initial

Explanation / Answer

Let the acid be represented as HA

HA <-----> H+ + A-
0.0158 0 0 (initial)
0.0158-x x x (at equilibrium)

[H+] = x + 10^-7 M
10^-7 M is due to water

pH =-log [H+]
6.42 = - log (x+10^-7)
x+10^-7 = 3.802*10^-7 M
x = 2.802*10^-7 M

Ka = x*x/(0.0158-x)
= (2.802*10^-7)*(2.802*10^-7) / (0.0158 - 2.802*10^-7)
= 4.969*10^-12

Answer: 4.969*10^-12

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