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

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?

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)

Answer 1

A)

we know that

for weak bases

[OH-] = sqrt ( Kb x C)

[OH-] = sqrt ( 6 x 10-6 x 0.175)

[OH-] = 1.0247 x 10-3

we know that

pOH = -log [OH-]

so

pOH = -log 1.0247 x 10-3

pOH = 2.99

so

the pOH is 2.99

B)

for weak bases

[OH-] = sqrt ( kb x C)

[OH-] = sqrt ( 6 x 10-6 x 0.325)

[OH-] = 1.3964 x 10-3

now

% ionization = [OH-] x 100 / [NX3]

% ionization = 1.3964 x 10-3 x 100 / 0.325

% ionization = 0.43

so

the % ionization is 0.43 %

C)

we know that

Ka x Kb = 10-14

so

Ka x 6 x 10-6 = 10-14

Ka = 1.666 x 10-9

now

pKa = -log Ka

so

pKa = -log 1.666 x 10-9

pKa = 8.778

so

the pKa for the reaction is 8.778