Exercise Number201 A buffer solution is prepared as described in (a) below. To a

Question

Exercise Number201 A buffer solution is prepared as described in (a) below. To a sample of this buffer solution is added a solution of a strong base as described in (b). (a) 100 mL of 0.11 M benzoic acid are mixed with 0.011 moles of sodium benzoate plus 100 mL of water are added. (b) 1.0 mL of 0.50 M KOH is added to 50 mL of the solution described in part (A). For each of these solutions calculate: (i) the molar concentration of benzoic acid. (i) the molar concentration of (iii) the pH of the resulting solution The K, of benzoic acid is 6.5 x 103 In the lab manual in the section of Making a Buffer to a Given pH" the buffer you will be required to prepare will have a pH of 4.47. Make note of this value and the prelab exercise number on your report sheet in the lab manual.

Explanation / Answer

a) total volume of solution = (100 + 100) = 200 ml.

so 100 ml of 0.11 M sodium benzoic acid = 0.100 * 0.11 = 0.011 mole.

molar concentration of benzoic acid = 0.011 * 1000 / 200 = 0.055 M

molar concentration sodium benzoate ion = 0.011 * 1000 / 200 = 0.055 M.

PKa = - log Ka = - log (6.5 * 10^-5) = 4.19

PH = PKa + log [salt]/ [acid]

PH = 4.19 + log (0.011 / 0.011)

PH = 4.19.

b) 50 ml of 0.011 M benzoic acid = 0.050 * 0.11 = 0.0055 mole.

50 ml of 0.011 M sodium benzoate = 0.050 * 0.011 = 0.0055 mole.

1.0 ml of 0.50 M KOH = 0.001 * 0.50 = 0.0005 mole.

molar concentration of benzoic acid = (0.0055 - 0.0005) * 1000 / (50 + 1) = 0.098 M

molar concentration of sodium benzoate = (0.0055 + 0.0005) * 1000 / 51 = 0.1176 M

PH = PKa + log [salt] / [ acid] = 4.19 + log (0.1176 / 0.098)

PH = 4.27

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