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!)

Answer 1

       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