If enough of a monoprotic acid is dissolved in water to produce a 0.0184 M solut

Question

If enough of a monoprotic acid is dissolved in water to produce a 0.0184 M solution with a pH of 6.62, what is the equilibrium constant, K_a, for the acid? K_a = Number 3.13 times 10^-12 You have not accounted for the autoprotolysis of water when calculating the K_a 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 K_a value. Start by writing four equations: the K_w expression the K_a expression the charge balance of the solution the material (mass) balance of the solution Use these four equations to develop an expression for K_a in terms of K_w, [H_3O^+], and the initial concentration of the acid, [HA]_initial. Recall that [H_3O^+] = 10^-pH

Explanation / Answer

pH = -log[H+] = 6.62

[H+] = 2.4 x 10^-7 M

[H+] from water = 1 x 10^-7 M

Ka = (2.4 x 10^-7 x 1 x 10^-7)^2/(0.0184 - 2.4 x 10^-7)

     = 3.13 x 10^-26

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