Question

140.0 mL of 0.25 M  HF with 230.0 mL of 0.32 M  NaF
The Ka of hydrofluoric acid is 6.8 x 10−4.

Express your answer using two decimal places.

185.0 mL of 0.11 M C2H5NH2 with 270.0 mL of 0.20 M C2H5NH3Cl

Express your answer using two decimal places.

Answer 1

Part-A:

The H-H equation has the form:

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

moles HF = 0.140 L X 0.25 mol/L = 0.035 mol HF

moles F- = 0.230 L X 0.32 mol/L = 0.0736 mol F-

Ka for HF = 6.8 X 10^-4. pKa = -log Ka = 3.17

pH = 3.17 + log (0.0736 / 0.035) = 3.49

pH = 3.49

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Part-B:

My text gives a Kb for ethylamine of 6.41 X 10^-4. Since Ka X Kb = 1 X 10^-14, Ka = 1.56 X 10^-11 and pKa = 10.81

moles base = 0.185 L X 0.11 mol/L = 0.02035 mol base

moles acid = 0.270 L X 0.20 mol/L = 0.054 mol acid

pH = 10.81 + log (0.02035 / 0.054 ) = 10.39

pH = 10.39

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140.0 mL of 0.25M HF with 230.0 mL of 0.32 M NaF The K, of hydrofluoric acid is 6.8 x 10 Express your answer using two decimal places. A O O ? TO ADD pH= 3.49 Submit Request Answer Part B 185.0 mL of 0.11 MC,H; NH, with 270.0 mL of 0.20 M C,H;NHCI Express your answer using two decimal places. 0 2 ? IVO AED pH = 10.39

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