CBSE Class 11-science Answered
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 | 06 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 | 08 Dec, 2010, 10:06: AM
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