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

Asked by arihant7 | 6th Dec, 2010, 08:06: PM

Expert Answer:

Dear Student

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

pOH = pKb - log(base/salt)

So log(base/salt) = pKb-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

Answered by  | 8th Dec, 2010, 10:06: AM

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