Map Sapling Learning macmellan larning If enough of a monoprotic acid is dissolv
ID: 550812 • Letter: M
Question
Map Sapling Learning macmellan larning If enough of a monoprotic acid is dissolved in water to produce a 0.0196 M solution with a pH of 6.69, what is the equilibrium constant, Ka, for the acid? Number K2.14x 1012 Incorrect. 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: 1) the Kw expression 2) the Ka expression 3) the charge balance of the solution 4) 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, [HAlnitial Recall that H30']= 10-pH Previous Give Up & View Solution Try Again Next ExitExplanation / Answer
we have below equation to be used:
pH = -log [H+]
6.69 = -log [H+]
log [H+] = -6.69
[H+] = 10^(-6.69)
[H+] = 2.042*10^-7 M
Lets write the dissociation equation of HA
HA -----> H+ + A-
0.0196 0 0
0.0196-x x x
Ka = [H+][A-]/[HA]
Ka = x*x/(c-x)
Ka = 2.042*10^-7*2.042*10^-7/(0.0196-2.042*10^-7)
Ka = 2.13*10^-12
Answer: 2.13*10^-12
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