Question

For each of the following solutions, calculate the initial pH and the final pH after adding 0.005 mol of NaOH.

290.0 mL of a buffer solution that is 0.315 molL−1 in C2H5NH2 and 0.295 molL−1 in C2H5NH3Cl

Answer 1

We know that,

pKb for C2H5NH2 = 3.37

[C2H5NH2] = 0.315 M

[C2H5NH3Cl] = 0.295 M

1) Initial pH

This is a basic buffer.

pOH of this basic buffer = pKb + log ([C2H5NH3Cl] / [C2H5NH2])

=> pOH = 3.37 + log (0.295 / 0.315) = 3.34

pH = 14 - pOH

=> pH = 14 - 3.34 = 10.66

2) Final pH

Moles of NaOH added = 0.005

Volume of solution = 290 mL = 0.29 L

=> [NaOH] = 0.005 / 0.29 = 0.0172 M

pOH after adding NaOH = pKb + log ([C2H5NH3Cl] - [NaOH] / [C2H5NH2] + [NaOH])

=> pOH = 3.37 + log (0.295 - 0.0172 / 0.315 + 0.0172) = 3.29

=> pH = 14 - 3.29 = 10.71