Consider a 2×10−2 M solution of the weak acid butanoic acid , for which pKa = 4.83 .
You can search for the structure of this compound online, although precise knowledge of the structure is not needed.
Your answers need to be within 5% of the correct answer to be considered correct. Don't round or you will fall outside the 5% margin of error.
Part A: Calculate [H+] (units in M)
Part B: Calculate the concentration of butanoate ions (units in M)
Part C: Calculate the concentration of butanoic acid in equilibrium. (units in M)
Part D: Calculate [OH−] (units in M)
Part E: Calculate the pH
pKa = 4.83
Ka = 10^-pKa
Ka = 1.48 x 10^-5
C = 2 x 10^-2 M
part A)
[H+] = sqrt (Ka x C)
[H+] = sqrt (1.48 x 10^-5 x 2 x 10^-2)
[H+] = 5.44 x 10^-4 M
part B)
concentration of butanoate ion = 5.44 x 10^-4 M
part C)
concentration of butanoic acid in equilibrium = 2 x 10^-2 - (5.44 x 10^-4)
= 0.0195 M
part D)
[H+] [OH- ] = 1.0 x 10^-14
5.44 x 10^-4 [OH-] = 1.0 x 10^-14
[OH-] = 1.84 x 10^-11 M
part E)
pH = -log [H+]
pH = -log (5.44 x 10^-4)
pH = 3.26
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