Question

What masses of ethylamine (K= 5.60 x 104) and ethylammonium chloride do you need to prepare 1.73 L of 10.548 buffer if the total concentration of the two components is 2.63 M pH 6th attempt See Periodic Table See Hint (0.5 point) Part 1 Ethylamine g Part 2 (0.5 point) Ethylammonium chloride g

Answer 1

Ans:

Kb of ethylamine = 5.60 x 10^-4

pKb of ethylamine = -logKb = -log(5.60 x 10^-4) = 3.2518

Total volume of the solution = 1.73 L

Total concentration of the two components = 2.63 M

pH of the solution = 10.548

pOH of the solution = 14-pH = 14-10.548 = 3.452

According to Henderson Hasselbalch equation, pOH of the buffer is given by,

pOH = pKb + log {[ethylammonium chloride]/[ethylamine]}

3.452 = 3.2518 + log {[ethylammonium chloride]/[ethylamine]}

[ethylammonium chloride]/[ethylamine] = 1.5856 --------------------> equation 1

Let n1 be the number of moles of ethylammonium chloride & n2 be the number of moles of ethylamine.

It is given that total concentration of the two components = 2.63 M = (n1 + n2)/1.73

So, (n1 + n2) = 4.5499 mol.

Substitution in equation 1, we will get

[ethylammonium chloride]/[ethylamine] = (n1/1.73)/(n2/1.73) = n1/n2 = 1.5856

n1/(4.5499-n1) = 1.5856

n1 = 2.79 = number of moles of ethylammonium chloride

n2 = 1.7598 mol = number of moles of ethylamine

Molar mass of ethylammonium chloride = 81.54 g/mol

Mass of 2.79 mol of ethylammonium chloride = 2.79 x 81.54 = 227.4966 g

Molar mass of ethylamine = 45.085 g/mol

Mass of 1.7598 mol of ethylamine = 1.7598 x 45.085 = 79.3406 g