If enough of a monoprotic acid is dissolved in water to produce a 0.0155 M solut

Question

If enough of a monoprotic acid is dissolved in water to produce a 0.0155 M solution with a pH of 6.43, what is the equilibrium constant, K_a, for the acid? K_a = 8.93 times 10^-12 Incorrect. 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

For a very weak acid...

we need to acount for Kw ionization of water H+ and OH-

so

pH = 1/2*(pKa - log(C))

in this case

6.43 = 1/2*(pKa - log(0.0155))

so

pKa= 6.43*2 + log(0.0155)

pKa = 11.0503

Ka = 10^-11.0503

Ka = 8.90635*10^-12

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