The [OH-] of a solution of 0.064 mol/L ammonia

Question

a) The [OH^-] of a solution of 0.064 mol/L ammonia is a.bcde mol/L

b) The [OH^-] of a solution of 3.05 mol/L sodium carbonate is a.bcd mol/L

Answer 1

a). NH₃ + H2O ==> NH₄OH ==> NH₄⁺ + OH^-

Kb NH₃ = 1.8x10^-5 (look it up)

1.8x10^-5 = [NH4+][OH-] / [NH3]

1.8x10^-5 = (x)(x) / 0.064

x² = 1.15x10-6

x = [OH-] = 1.07x10^-3 M

a = 0

b = 0

c = 0

d = 1

e = 1

b). Na₂CO3 + H₂O ==> 2NaOH + CO₂

In 1 liter: 3.05 mol Na2CO3 x 2 mol NaOH / mol Na2CO3 = 6.10 mols NaOH = 6.10 mols OH-

a = 6

b = 1

c = 0

d = 0

You should mention to your teacher that reporting the answer to 4 sig. figs. is incorrect, as is the case for both parts of this question.