What is the pH of a buffer made by combining 55.0 ml of 0.450 M HCN...

What is the pH of a buffer made by combining 55.0 ml of 0.450 M HCN and 45.0 ml of 0.250 M NaCN?
(Ka of HCN = 4.4 x 10^-10) Use the Henderson Hasselbach equation, and prove why you can use it.

Expert Solution

this is a buffer, as stated, use HH equation

pH = pKa + log(NaCH/HCN)

recalculate mmoles in solutino

mmol of NaCN = MV = 45*0.25 = 11.25

mmol of HCN = MV = 0.45*55 = 24.75

pKa 0 -log(HACN) - log(4.4*10^-10) = 9.3565

pH = 9.3565 + log(11.25/24.75)

pH = 9.0140

queen_honey_blossom answered 2 years ago

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