1) jake wants to make a pH 5.5 pyridine buffer by adding concentrated HCl to 500ml...

1) jake wants to make a pH 5.5 pyridine buffer by adding concentrated HCl to 500ml of a 0.05M pyridine solution. if the Ka is 1.26*10^-5, how many ml of 0.5M HCl should be added?

2) Describe the preparation of 2.0 liters of 0.4M HCl by measuring out a calculated volume. You begin with a 28% w/w solution of HCl (FW=36.5) that is 1.15 g/ml

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Expert Solution

pka of pyridine = -log(1.26*10^-5) = 4.9

as pyridine is a base, pkb = 14-4.9 = 9.1

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

no of mole of pyridine = M*V = 500*0.05 = 25 mmol

no of mole of HCl must add = x mmol

5.5 = 14-(9.1+log(x/(25-x))

x = 5.02

no of mole of HCl must add = x = 5.02 mmol

volume of HCl must add = n/M = 5.02/0.5 = 10.04 ml

Molarity(M) = w/mwt*(1/v in L)

Mwt = molarmass of HCl = 36.5 g/mol

w = mass of HCl required = ?

v = volume of solution = 2 L

Molarity = 0.4 M

0.4 = (w/36.5)*(1/2)

w = 29.2

w/w% = 28% of HCl solution. means 28 grams HCl present in 100 g solution.

required solution = 29.2*100/28 = 104.3 grams

d = density of solution = 1.15 g/ml

volume of 28% of HCl solution required = m/d = 104.3/1.15 = 90.7 ml

take 90. 7 ml of 28% of HCl solution and add water upto 2 L

queen_honey_blossom answered 1 year ago

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