Question

A solution of household ammonia is 3.0 % NH₃(aq) and has a density of .98 g/ml. What is the molar concentration of the NH₃(aq)? Given that the K_b for NH₃(aq) = 1.8x10^‒5, what is the predicted pH of this solution?

Thank you in advance

Answer 1

1)

Let volume of solution be 1 L

volume , V = 1 L

= 1*10^3 mL

density, d = 0.98 g/mL

use:

mass = density * volume

= 0.98 g/mL *1*10^3 mL

= 9.8*10^2 g

This is mass of solution

mass of NH3 = 3.0 % of mass of solution

= 3.0*980.0/100

= 29.4 g

Molar mass of NH3,

MM = 1*MM(N) + 3*MM(H)

= 1*14.01 + 3*1.008

= 17.034 g/mol

mass(NH3)= 29.4 g

use:

number of mol of NH3,

n = mass of NH3/molar mass of NH3

=(29.4 g)/(17.03 g/mol)

= 1.726 mol

volume , V = 1 L

use:

Molarity,

M = number of mol / volume in L

= 1.726/1

= 1.726 M

Answer: 1.73 M

2)

NH3 dissociates as:

NH3 +H2O -----> NH4+ + OH-

1.726 0 0

1.726-x x x

Kb = [NH4+][OH-]/[NH3]

Kb = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Kb = x*x/(c)

so, x = sqrt (Kb*c)

x = sqrt ((1.8*10^-5)*1.726) = 5.574*10^-3

since c is much greater than x, our assumption is correct

so, x = 5.574*10^-3 M

So, [OH-] = x = 5.574*10^-3 M

use:

pOH = -log [OH-]

= -log (5.574*10^-3)

= 2.2538

use:

PH = 14 - pOH

= 14 - 2.2538

= 11.7462

Answer: 11.75