Question

Consider a 0.20 M NH4Cl aqueous solution. Rank the following species in the solution from highest to lowest concentration.

NH3, Cl-, H2O, NH4+, OH-, H3O+

Consider a 0.10 M aqueous solution of sodium nitrite, NaNO2. Rank the following species in the solution from highest to lowest concentration.  

OH-, HNO2, NO2-, H3O+, H2O, Na+

The acetate ion is the conjugate base of acetic acid. The Ka of acetic acid is 1.76×10–5. What is the pH of a 0.100 M solution of sodium acetate?

Answer 1

1) Consider a 0.20 M NH₄Cl aqueous solution. Rank the following species in the solution from highest to lowest concentration.

H₂O > Cl^- > NH₄⁺ > H₃O⁺ > NH₃ > OH^-

2) Consider a 0.10 M aqueous solution of sodium nitrite, NaNO₂. Rank the following species in the solution from highest to lowest concentration.  

H₂O > Na⁺> NO₂- > OH^- HNO₂ > H₃O⁺

3)

We have,

Kb = (1.0 x 10^-14)/Ka = (1.0 x 10^-14)/1.76 x 10^-5 = 5.682 x10^-10

CH₃COO^-  + H₂O       CH₃COOH +   OH^-

0.1                                              0                    0

0.1-x                                          x                     x

Kb = [CH₃COOH][OH^-]/[CH₃COO^-]

Kb = x²/(0.1-x)

x can be ignored as compared to 0.1 M

So, above expression becomes

Therefore,

x² = 5.682 x 10^-10 x 0.1

x² = 5.682 x 10^-11

x = 7.53 x 10^-6

pOH = -log [OH^-] = -log (7.53 x 10^-6) = 5.12

PH = 14 – pOH = 14 - 5.12 = 8.87

Answer: 8.87