Question

1. 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)    1.04 ✕ 10⁻¹M  
(b)    1.04 ✕ 10⁻²M
(c)    1.04 ✕ 10⁻¹²M

2.How many milliliters of 0.248 M HNO₃ should be added to 217 mL of 0.00662 M 2,2'-bipyridine hydrochloride to give a pH of 4.18? (Assume pK_a = 4.34 and K_w = 1.01 ✕ 10⁻¹⁴.)

Answer 1

The association reaction of sodium butanoate is

-----------CH3CH2CH2COO^-(aq) + H2O --- > CH3CH2CH2COOH + OH^-(aq), Kb = Kw / Ka = 1.01x10^-14 / 1.52x10^-5

Init.conc(M): 1.04x10^-1 ------------------------------------ 0 -------------------------- 0

eqm.conc(M):(1.04x10^-1 - ) --------------------------- -------------------------

We calculate Kb

=> Kb = 1.01x10^-14 / 1.52x10^-5 = 6.645x10^-10

^{then we can calculate ;}

a) =>  6.645x10^-10 = ² / (1.04x10^-1 - )

=> ² + 6.645x10^-10x - 6.911x10^-11 = 0

=>   = 8.31 x 10^-6

b)

=>  6.645x10^-10 = ² / (1.04 ✕ 10⁻² - )

=> ² + 6.645x10^-10x - 6.908x10^-12 = 0

=>   = 2.6283 x 10^-6

c)

=>  6.645x10^-10 = ² / (1.04x10^-12 - )

=> ² + 6.645x10^-10x - 6.911x10^-22 = 0

=>   = 4.3879 x 10^-19 (discriminant)

2) 2,2'-bipyridine hydrochloride is a salt and so it will not react with HNO₃. So the pH that is expected will come entirely from HNO₃.

2C₅H₅N → (C₅H₄N)₂ + H₂ (is a weak base)

C10H8N2 + H3O⁺ ---------------> C10H8N2H⁺ + H2O

if pH = 4.18

pH = -log [H⁺]

[H⁺] = 10^-pH

[H⁺] = 10^-4.18

[H⁺] = 6.6 x 10^-5M is the final concentration required

6.6 x 10^-5M means 6.6 x 10^-5 moles in 1L

-----------

then;

pK_a = 4.34

K_w = 1.01 ✕ 10⁻¹⁴

KaKb = Kw

Kw = [H+][OH-]

pH = pKa - log [A-]/[HA]

pKa = -log Ka

4.34 = -log Ka

Ka = 4.5 x 10^-5 = [H3O⁺][C10H8N2] / [C10H8N2H⁺]

-------

4.5 x 10^-5 / 6.6 x 10^-5 = [C10H8N2] / [C10H8N2H⁺]

[C10H8N2] / [C10H8N2H⁺] = 6.81 x 10^-11

initial mol C10H8N2 = (217mL)(0.00662 mol/1000ml) = 0.001436 mol

--------

if x = mol C10H8N2H = mol HNO3

6.81 x 10^-11 = (0.001436 - x) / x

x = 1.43 x 10^-3

------

finally we can calculate the milimeters of 0.248 mol of HNO3:

1.43 x 10^-3 mol HNO3 x 1000ml / 0.248 mol HNO₃ = 5.79 ml of HNO₃

Hope it helps!