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 =
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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