Consider the balanced reaction of acetylsalicylic acid (HA) to produce benzoic a
ID: 104105 • Letter: C
Question
Consider the balanced reaction of acetylsalicylic acid (HA) to produce benzoic acid (HB).
HA(aq) + B^1-(aq) <=> A^1-(aq) + HB(aq)
At 25°C the equilibrium constant of this reaction is 5.71. Assuming one begins with 0.3740-M of both HA and B^1- (nothing else), (a) determine the equilibrium amounts of [HA] and [HB] at 25°C. Hint: a quadratic should not be required. Retain four decimal digits in the reaction table and answer.
Let us now consider a different equilibrium mixture for the same reaction and temperature as given in the previous problem, so that K is still = 5.71, viz:
HA(aq) + B^1-(aq) ó A^1-(aq) + HB(aq)
Given the initial equilibrium amounts of all species as:
[HA] = 0.0247-M [B^1-] = 0.0299-M [A^1-] = 0.0488-M [HB] = 0.0864-M
Determine: (b) how many mol/liter (M) of A^1- must be removed so as to raise the equilibrium concentration of HB to a value of 0.0900-M (i.e. by +0.0036-M) and (c) the final (new) equilibrium concentration of [A^1-]. Retain four decimal digits.
(The answers should be a: [HA]25 = 0.1104M [HB]25 = 0.2636M; b:0.0172M of A^1- removed; c: 0.0352 = [A^1-] eq but I dont know how to get these answers - please explain!)
Explanation / Answer
HA(aq) + B^1-(aq) <=====> A^1-(aq) + HB(aq)
Intial 0.374 M 0.374 M 0 M 0 M
change - x -x +x +x
equili 0.374-x 0.384-x x x
K = [A^1-][HB]/[B^1-][HA]
5.71 = (x*x)/(0.374-x)^2
x = 0.2636
at equilibrium,
[A^1-] = x = 0.2636 M
[HB] = x = 0.2636 M
[B^1-] = 0.374-X = 0.374-0.2636 = 0.1104 M
[HA] = 0.374-X = 0.374-0.2636 = 0.1104 M
B)
FROM the given data,
K = [A^1-][HB]/[B^1-][HA]
5.71 = ((0.0488-x)*(0.09))/((0.0299-x)*(0.0247-x))
[A^1-] must removed = x = 0.0396 M
c) final equilibrium [A^1-] = 0.0488-0.0396 = 0.0092 M