Question

For each of the following solutions, calculate the initial pH and the final pH after adding 0.005 mol of NaOH290.0 mL of a buffer solution that is 0.315 molL−1 in C2H5NH2 and 0.295 molL−1molL−1 in C2H5NH3Cl

Answer 1

moles of C2H5NH3Cl = molarity x volume = 0.295 M x 0.29 L = 0.085 mol

moles of C2H5NH2 = molarity x volume = 0.315 M x 0.29 L = 0.091 mol

[C2H5NH3Cl] = 0.085 mol

[C2H5NH2] = 0.091 mol

Kb of C2H5NH2 = 5.6 x 10^-4

Initial pH

pOH = pKb + log ([salt] / [base])

pOH = pKb + log ([C2H5NH3Cl] / [C2H5NH2])

pOH = -logKb + log ([C2H5NH3Cl] / [C2H5NH2])

= - log (5.6×10⁻⁴) + log (0.085 mol / 0.091 mol)

= 3.22

pOH = 3.22

Then,

pH = 14 -pOH = 14 - 3.22 = 10.78

Therefore,

Initial pH = 10.78

pH after addition of 0.005 mol NaOH:

C2H5NH3Cl + NaOH --------------> C2H5NH2 + NaCl + H2O

0.085 mol 0.005 mol 0

------------------------------------------------------------------------------------

0.085-0.005 0 0.005 mol

= 0.08 mol

Then,

[C2H5NH3Cl] = 0.08 mol

[C2H5NH2] = 0.091 mol + 0.005 mol = 0.96 mol

pOH = -logKb + log ([C2H5NH3Cl] / [C2H5NH2])

= - log (5.6×10⁻⁴) + log (0.08 mol / 0.096 mol)

= 3.17

pOH = 3.17

Then,

pH = 14 -pOH = 14 - 3.17 = 10.83

Therefore,

Final pH = 10.83