Question

How many milliliters of 0.246 M HCl should be added to 213 mL of 0.00666 M ethylamine to give a pH of 10.52?

Answer 1

Kb of ethylamine = 6.4*10^-4
pKb = -log Kb
= -log (6.4*10^-4)
= 3.194

pH = 10.52
pOH = 14 - 10.52 = 3.48

USE:
pOH = pKb + log {[C2H5NH3+]/[C2H5NH2]}
3.48 = 3.194 + log {[C2H5NH3+]/[C2H5NH2]}
[C2H5NH3+]/[C2H5NH2] = 1.932

let V ml of HCl is added
then, after reaction between ethylamine and HCl
mol of C2H5NH3+ = 0.246*V mmol
mol of C2H5NH2 = 0.00666*213 - 0.246*V

[C2H5NH3+]/[C2H5NH2] = 1.932
(0.246*V) / (0.00666*213 - 0.246*V ) = 1.932
(0.246*V) / (1.419 - 0.246*V ) = 1.932
0.246*V = 2.741 - 0.475*V
V = 3.802 mL

Answer: 3.802 mL