Question

A 23.9 mL sample of 0.345 M ethylamine, C2H5NH2, is titrated with 0.346 M hydrobromic acid. After adding 9.17 mL of hydrobromic acid, the pH is -blank-

Answer 1

no. of mole = molarity volume of solution in liter

no.of mole of C₂H₅NH₂  = 0.345 M 0.0239 L = 0.0082455 mole

no. of mole of HBr = 0.346 M 0.00917 L = 0.00317282 mole

netrilization reaction between C₂H₅NH₂ and HBr take place 0.00317282 mole of HBr neutrilize 0.00317282 mole of C₂H₅NH₂

mole of C₂H₅NH₂ remain in solution = 0.0082455 - 0.00317282 = 0.00507268 mole

total volume of solution = 23.9 + 9.17 = 33.07 ml = 0.03307 liter

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

Molarty of C₂H₅NH₂ = 0.00507268/ 0.03307 = 0.153 M

C₂H₅NH₂ dissociate as

C₂H₅NH₂ + H₂O   C₂H₅NH₃⁺ + OH^-

K_b = [C₂H₅NH₃⁺ ] [OH^-] / [C₂H₅NH₂]

ethylmine is monobesic base therefore

[C₂H₅NH₃⁺ ] = [OH^-] = x

K_b = [x][x] /  [C₂H₅NH₂]

K_b =[x]² /  [C₂H₅NH₂]

[x]² = K_b  [C₂H₅NH₂]

K_b of C₂H₅NH₂ = 5.6 10^-4

[x]² = 5.6 10^-4 0.153 = 8.568 10^-5

[x] = 0.009256 M

[C₂H₅NH₃⁺] = [OH^-] = x = 0.009256 M

pOH = -log[OH^-] = -log(0.009256) = 2.03

pH = 14 - pOH = 14 - 2.03 = 11.97

pH = 11.97