Question

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

140.0 mL of 0.27 M  HF with 225.0 mL of 0.30 M  NaF

175.0 mL of 0.12 M C2H5NH2 with 265.0 mL of 0.20 M C2H5NH3Cl

Express your answer using two decimal places.

Answer 1

pH of the solution for an acid is calculated using Henderson-Hasselbach equation given by:

pH = pKa+log[acid]/[conjugate base]

pH of the solution for a base is calculated using Henderson-Hasselbach equation given by:

pOH = pKb+log[base]/[conjugate acid]

pH of the solution for the mixture 140.0 mL of 0.27M HF with 225.0 mL of 0.30M NaF is calculated as follows:

Number of moles of HF = 0.140L0.27M = 0.038

Number of moles of F^- = 0.225L0.30M = 0.068

Total volume of the mixture = 0.140L+0.225L = 0.365L

[HF] = 0.0380.365 = 0.104M

[F^-] = 0.0680.365 = 0.186M

pKa for HF is 3.18

Therefore, pH = 3.18+log[F-][HF] = 3.18+log0.1860.104

pH = 3.43

pH of the solution for the mixture 175.0 mL of 0.12M C₂H₅NH₂ with 265.0 mL of 0.20M C₂H₅NH₃Cl is calculated as follows:

Number of moles of C₂H₅NH₂ = 0.175L0.12M = 0.021

Number of moles of C₂H₅NH₃Cl  = 0.265L0.20M = 0.053

Total volume of the mixture = 0.175L+0.265L = 0.440L

[C₂H₅NH₂] = 0.0210.440 = 0.048M

[C₂H₅NH₃Cl ] = 0.0530.440 = 0.120M

pKb for C₂H₅NH₂ is 3.19

Therefore, pOH = 3.19+log[C₂H₅NH₃⁺ ][C₂H₅NH₂] = 3.19+log0.1200.048

pOH = 3.59

Therefore, pH = 14-pOH = 14-3.59

pH = 10.41