Consider a 2×10 2 M solution of the weak acid butanoic acid , for which p K a =

ID: 548831 • Letter: C

Question

Consider a 2×102 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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