Question

A buffer is prepared by mixing 48.3 g of NH3 and 48.3 g of NH4Cl in...

A buffer is prepared by mixing 48.3 g of NH3 and 48.3 g of NH4Cl in 0.600 L of solution. What is the pH of this buffer, and what will the pH change to if 7.22 g of HCl is then added to the mixture?

pH of this buffer=

The pH change =

Homework Answers

Answer #1


pH of basic buffer = 14-(pkb + log(salt/base))

pkb of NH3 = -log(1.8*10^-5) = 4.74

no of mol of salt (NH4Cl) = 48.3/53.49 = 0.903 mole

no of mol of base(NH3) = 48.3/17 = 2.84 mole

pH = 14 - (4.74 + log(0.903/2.84))

initial pH   = 9.76

after HCl added

no of mole of HCl = 7.22/36.5 = 0.198 mole

pH = 14-(pkb + log(salt+HCl/base-HCl))

    = 14 - (4.74+log((0.903+0.198)/(2.84-0.198))

    = 9.64

pH change = 9.76 - 9.64 = 0.12

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