Question

A 1.100 L sample of HCl gas at 25∘C and 732.0 mm Hg was absorbed completely in an aqueous solution that contained 6.900 g of Na2CO3 and 265.0 g of water.

What is the pH of the solution?

What is the freezing point of the solution?

What is the vapor pressure of the solution? (The vapor pressure of pure water at 25 ∘C is 23.76 mm Hg.)

Answer 1

The number of mols of HCl = n = PV / RT = (732/760) atm*1.100 L /(0.0821*298)

                                             = 0.0433 mol

Thus, concentration of the solution = 0.0433 mol / 0.265 L = 0.163 mol/L

pH = -log [H⁺] = -log (0.163) = 0.786

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Depression in freezing point = K_f* m

                                         = (1.86⁰C) (6.9 g / 106 g/mol) (1000 / 265)

                                         = 0.45⁰C

Hence - Freezing point of the substance = -0.45⁰C

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