Calculate the amount of (NH4)2SO4 which must be added to 500 ml of 0.2(M) NH3 to yield a solution of pH=9.35 pkb of NH4OH=4.74

Dear Student

pH = 9.35 => pOH = 14-9.35 = 4.65

pOH = pK_b - log(base/salt)

So log(base/salt) = pK_b-pOH = 4.74-4.65 = 0.09

So base/salt = 1.23

Number of moles of NH₃ = 0.2*0.5 = 0.1 mole

So,

0.1/ salt = 1.23

salt conc. = 0.08

0.08 mole of ammonium ion = 0.04 mole [NH₄]₂SO₄

Reagrds

Topper team