Question

1) a) How would you make 600 ml of a 0.25 M ammonia and 0.20 M ammonium nitrate from a bottle of 6.0 M ammonia (aq) and a bottle of solid ammonium nitrate.

grams of ammonium nitrate?
   mL of ammonia?
mL of H2O?

   b) Calculate pH (Kb = 1.8 x 10^-5)

2) How many mL of Acid (12 M HCl) or base (12 M NaOH) would you need to shift the pH 0.5 increments down?

Answer 1

1)

a) V= 600 ml,

Conc. of NH₃ =0.25 M, Conc. of NH₄NO₃ =0.20 M,

Molar mass of NH₃= 17.0319 g/mol, density of NH3= 0.73 g/L

Molar mass of NH₄NO₃ = 80.043 g/mol, density of NH₄NO₃= 1.72 g/ml

Now, first calculate number of moles of each solute in final solution:

SO,

Moles of NH₃ = 0.25 M * 0.600 L = 0.150 mol

Moles of NH₄NO₃ = 0.20 M* 0.600 L = 0.120 mol

Volume of 6 M solution required to make 0.25 M NH₃ is: 0.150 mol / 6 mol/L = 0.025 L = 25 ml

Mass of NH₄NO₃ required to make 0.20M NH₄NO₃ (0.120 mol NH4NO3 ) is

Mass of 0.20M NH₄NO₃  (0.120 mol NH₄NO₃ ) = moles*molar mass = 0.120*80.043 = 9.61 g

Volume of NH₄NO₃ = Mass /density = 9.61 g / 1.72 g/ml = 5.58 ml

Volume of H₂O required = 600 -Volume of NH₃ -Volume of NH₄NO₃

Volume of H₂O required = 600 - 25 - 5.58 = 569.42 ml   

b) pH of solution:

K_b= 1.8 x 10^-5

NH₄+ + H₂O = NH₃ + H₃O+

Now calculate Ka value

10-14 = K_a*  K_b

K_a = 10-14 / 1.8 x 10^-5

K_a = 5.56*10^-10

Now calculate the [H+]

K_a = [NH₃][H+]/[NH₄+]

Note: [NH₃]=[H+]

K_a= [H+]²/[NH4+]

5.56*10^-10 =  [H+]² / 0.20

[H+]² = 1.10*10^-10

[H+]=1.05*10^-5

pH= -log [H+]= -log1.05*10^-5 = 4.98

pH= 4.98