Question

Calculate the pH of the solution that results from each of the following mixtures.

A. 160.0 mL of 0.25 M HF with 225.0 mL of 0.32 M NaF

B. 185.0 mL of 0.11 M C2H5NH2 with 280.0 mL of 0.20 M C2H5NH3Cl

Answer 1

A) mixture of HF and NAF act as acidic buffer

pH = pKa + log [NAF] / [HF]

pKa of HF = 3.17

[NaF] = (225 x 0.32 / 160+225) = 0.187 M

[HF] = (160 x 0.25 / 160 + 225) = 0.104 M

pH = 3.17 + log [0.187] / [0.104]

pH = 3.42

B) mixture of C2H5NH2 and C2H5NH3Cl act as basic buffer

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

pKb of C2H5NH2 =3.19

[C2H5NH2] = (185 x 0.11 / 185 + 280) = 0.044 M

[C2H5NH3Cl] = (280 x 0.20 / 465) = 0.120 M

pOH = 3.19 + log [0.120] / [0.044]

pOH = 3.62

pH = 14 - 3.62

pH = 10.38