Question

In: Chemistry

Calculate the pH of the solution that results from each of the following mixtures. 140.0 mL...

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.

Solutions

Expert Solution

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


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