Question

A 20.8 mL sample of 0.376 M ethylamine, C2H5NH2, is titrated with 0.320 M hydrochloric acid. After adding 9.41 mL of hydrochloric acid, the pH is?

kb for ethylamine is 4.3 *10^-4

Answer 1

no. of mole = Molarity volume of solution in liter

no. of mole of C₂H₅NH₂  = 0.376 0.0208 = 0.0078208 mole

no. of mole of HCl added = 0.320 0.00941 = 0.0030112 mole

nutrilization reaction between ethylamine and HCl take place in equimolar proportion therefore 0.0030112 mole of HCl react with 0.0030112 mole of ethylamine.

mole of ethylamine remain in solution = 0.0078208 - 0.0030112 = 0.0048096 mole

volume of solution = 20.8 ml + 9.41 ml = 30.21 ml = 0.03021 liter

Molarity = no. of mole / volume of solution in liter

Molarity of ethyl amine after addition of HCl = 0.0048096 / 0.03021 = 0.159 M

ethylamin dissociated as

C₂H₅NH₂  + H₂O     C₂H₅NH₃⁺  + OH^-

Weak acid dissociates as AH ⇌ A- + H+

K_b = [C₂H₅NH₃⁺][OH^-] / [C₂H₅NH₂]

but  [C₂H₅NH₃⁺] = [OH^-] = x

K_b = [x][x] / [C₂H₅NH₂]

[x]² = K_b   [C₂H₅NH₂]

[x]²  = 4.3 10^-4 0.159

[x]²  = 6.837 10^-5

[x] = 0.0082686 M

Concentration of OH^- = 0.0082686 M

pOH = - log[OH^-]

pOH = - log (0.0082686)

pOH = 2.083

pH = 14 - pOH = 14 - 2.083 = 11.917

The pH of solution = 11.917