Question

Part A

If Kb for NX3 is 6.0×10⁻⁶, what is the pOH of a 0.175 M aqueous solution of NX3?

Express your answer numerically.

Part B

If Kb for NX3 is 6.0×10⁻⁶, what is the percent ionization of a 0.325 M aqueous solution of NX3?

Express your answer numerically to three significant figures.

Part C

If Kb for NX3 is 6.0×10⁻⁶ , what is the the pKa for the following reaction?

HNX3+(aq)+H2O(l)⇌NX3(aq)+H3O+(aq)

Express your answer numerically to two decimal places.

Answer 1

PART A

Ionisation equillibrium of NX3 is

NX3(aq) + H2O(l) ---------> HNX3⁺(aq) + OH^-(aq)

K_{​​​​​​b}= [HNX3⁺][OH^-]/[NX3]

at equillibrium ,

[NX3] = 0.175 - x

[HNX3] = x

[OH^-] = x

so,

x²/(0.175 - x) = 6.0 ×10^-6

solving for x

x = 0.001022

[OH^-] = 0.001022M

pOH = -log[OH^-]

pOH = -log(0.001022)

pOH = 2.99

PART B

Kb = [HNX3][OH^-]/[NX3] = 6.0×10^-6

at equillibrium

[NX3] = 0.325 - x

[HNX3] = x

[OH^-] = x

so,

x²/(0.325 - x) = 6.0×10^-6

solving for x

x = 0.001393

[HNX3⁺] = 0.001393M

% ionization = ([HNX3⁺]/[NX3]_iniial) × 100

% ionization =( 0.001393M/0.325M)× 100

% ionization = 0.43%

PART C

HNX3⁺ (aq) + H2O(l) <-------> NX3(aq) + H3O⁺(aq)

K_a = [NX3][H3O⁺] /[HNX3⁺]

K_a = K_w/K_b

where , K_w is ionic product of water, 1.00×10^-14

K_a = 1.00 ×10^-14/6.0 ×10^-6 = 1.67×10^-9

pKa = -logKa

pKa of HNX3⁺ = -log(1.67 ×10^-9)

= 8.78