In: Chemistry
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
1)
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
2)
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