Question

For 470.0 mL of a buffer solution that is 0.160 M in CH3CH2NH2 and 0.145 M in CH3CH2NH3Cl, calculate the initial pH and the final pH after adding 0.010 mol ofHCl.

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Answer 1

For 470.0 mL of a buffer solution that is 0.160 M in CH3CH2NH2 and 0.145 M in CH3CH2NH3Cl,

calculate the initial pH and the final pH after adding 0.010 mol ofHCl.

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(1)
Initial pH of basic buffer from Henderson-Hasselbach equation:
pOH = pKb + Log([Salt]/[Base])
[Salt] = [CH3CH2NH3Cl] = 0.145 M
[Base] = [CH3CH2NH2] = 0.160 M
pKb of base CH3CH2NH2 from table, pKb = 3.25

pOH = pKb + Log([Salt]/[Base]) = 3.25 + Log(0.145/0.16) = 3.2
pH = 14 -pOH = 14-3.2=10.8
(2)
pH after adding 0.010 mol of HCl

The following reaction will go to completion to form the weak acid, CH3CH2NH3+
CH3CH2NH2 + H+ -> CH3CH2NH3+
Stoichiometry calculation:

Before reaction:470.0 mL of a buffer solution that is 0.160 M in CH3CH2NH2 and 0.145 M in CH3CH2NH3Cl
[CH3CH2NH2] = 0.47*0.16 = 0.0752 mol
[H+] = [HCl] = 0.01 mol
[CH3CH2NH3+] = 0.47*0.145 = 0.06815 mol

After reaction:
Stoichiometry reaction CH3CH2NH2 + H+ -> CH3CH2NH3+ will be controlled by
limiting reactant present as [H+] = 0.01 mol

[CH3CH2NH2] = 0.0752 - 0.01 = 0.0652 mol
[H+] = [HCl] = 0.01-0.01 = 0 mol
[CH3CH2NH3+] = 0.06815 + 0.01 = 0.07815 mol

Molar concentrations in 0.47 L of solution:
[Base] = [CH3CH2NH2] = 0.0652/0.47 = 0.14 M
[Salt] = [CH3CH2NH3+] = 0.07815/0.47 = 0.166 M

from Henderson-Hasselbach equation:
pOH = pKb + Log([Salt]/[Base]) = 3.25 + Log(0.166/0.14) = 3.32
pH = 14 -pOH = 14-3.32=10.7