Question

Quick Solve Please

2.        15.0 mL of 0.50 M NaOH is added to a 100.-mL sample of 0.442 M NH₃ (K_b for NH₃ = 1.8 ´ 10^–5). What is the equilibrium concentration of NH₄⁺ ions? (3%)

3.        What is the percent dissociation of HNO₂ when 0.074 g of sodium nitrite is added to 115.0 mL of a 0.068 M HNO₂ solution? K_a for HNO₂ is 4.0 × 10^–4. (4%)

Answer 1

We know that

                                NH3 + H2O --> NH4+ + OH-

Moles of NaOH added = 0.0075

Kb = [NH4+] [OH- / [NH3] = 1.8 X 10^5

Initial mole of NH3 = molarity x volume = 0.442 X 0.1 = 0.0442 moles

                                   NH3 + H2O --> NH4+ + OH-

Initialy                      0.0442                       0      0.0075

let x moles of NH3 used in the reaction then at equilibrium

[OH-] = (x + 00075) / 0.115

[NH4+] =x / 0.115

[NH3] = 0.0442 - x / 0.115

Kb = 1.8 X-5= x (x+0.0075) /( 0.0442 -x ) 0.115

We can ignore x as it is << 1

so 1.8 X 10^-6 X 0.115 = x ^2 + 0.0075x / 0.0442

0.0091X 10^-6 = x^2 + 0.0075x

x^2 + 0.0075x -0.0091 X 10^-6 = 0

On solving

x = 0.0000119 M = [NH4+]