Question

Calculate the fraction of association (α) for the following concentrations of sodium butanoate. (Assume K_a = 1.52 ✕ 10⁻⁵ and K_w = 1.01 ✕ 10⁻¹⁴. Enter unrounded values.)

(a)    1.18 ✕ 10⁻¹ M


(b)    1.18 ✕ 10⁻² M


(c)    1.18 ✕ 10⁻¹² M

Answer 1

                              CH3CH2CH2COO^- ---> CH3CH2CH2COOH               +             OH^-

Initial:                   C                                           0                                                          0            

Final:                    (1-α)C                                  αC                                                       αC

K_b = K_w/K_a = 6.64*10^-10 = [CH3CH2CH2COOH][ OH^-]/[ CH3CH2CH2COO^-]

= αC* αC/(1-α)C

= α²C/(1-α)

or, α²C + αK_b - K_b = 0

or, α =[-K_b + √ (K_b² + 4K_bC)]/2C

1. C = 0.118 M

α =[-K_b + √ (K_b² + 4K_bC)]/2C

= [-(6.64*10^-10 ) + √ {(6.64*10^-10 )² + 4(6.64*10^-10 )0.118}] / (2*0.118)

= 7.5*10^-5

2. C = 0.0118 M

α =[-K_b + √ (K_b² + 4K_bC)]/2C

= [-(6.64*10^-10 ) + √ {(6.64*10^-10 )² + 4(6.64*10^-10 )0.0118}] / (2*0.0118)

= 2.37*10^-4

3. C = 1.18 *10^-12 M

α =[-K_b + √ (K_b² + 4K_bC)]/2C

= [-(6.64*10^-10 ) + √ {(6.64*10^-10 )² + 4(6.64*10^-10 )( 1.18 *10^-12 )}] / [2*(1.18 *10^-12)]

= 1.19

[NB: value of α should be less than 1. But it exceeds 1 in the third question. This is quite odd.]