Question

For 550.0 mL of a buffer solution that is 0.185 M in CH3CH2NH2 and 0.175 M in CH3CH2NH3Cl, calculate the initial pH and the final pH after adding 0.020 mol of HCl. Express your answers using two decimal places separated by a comma.

Answer 1

V = 550 ml of buffer

M = 0.185 M CH3CH2NH2

M = 0.175 M CH3CH2NH3Cl

calculate pH when adding 0.02 mol of HCl

In solution:

CH3CH2NH2 +H2O <--> CH3CH2NH3+and OH-

pKb = 3.3

CH3CH2NH3Cl ---> CH3CH2NH3+ and Cl-

There is common ion, therefore expect a buffer (basic)

When adding 0.02 mol of HCl...

CH3CH2NH2 + HCl ---> CH3CH2NH3+ and H2O + Cl-

Find moles of CH3CH2NH2 = M1*V1 = 0.185*550 = 101.75 mmol of CH3CH2NH2 initially

CH3CH2NH2 after reaction = 0.10175 - 0.02 = 0.08175 mol of CH3CH2NH2 left

Find moles of CH3CH2NH3+ = M2*V2 = 0.175*550 = 96.25 mmol of CH3CH2NH3+ initially

CH3CH2NH2 after reaction = 0.09625 + 0.02 = 0.11625 mol of CH3CH2NH3+ remain

For initial pH

Substitute in pOH equation

pOH = pKb + log(HB+/B)

pOH = 3.3 + log(0.10175 /0.09625 ) = 3.32

pH = 14-pOH = 14-3.32= 10.68

pH final = 10.68

For final pH

Substitute in pOH equation

pOH = pKb + log(HB+/B)

pOH = 3.3 + log(0.11625/0.08175 ) = 3.45

pH = 14-pOH = 14-3.45 = 10.55

pH final = 10.55