What is the pH of a solution of 0.15 M formic acid?

One litre of a buffer solution was prepared by dissolving 0.005 mole of formic acid and?

  • One litre of a buffer solution was prepared by dissolving 0.005 mole of formic acid and 0.0070 mole of sodium formate in water. Ka for formic acid is 1.8 x 10-4. (a) Calculate the pH of the solution. (b) If this solution was diluted to ten times its volume what would the pH be?

  • Answer:

    a) 3.89 b) 3.89

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(a) First of all calculate the molarity of the formic acid and the formate ion produced. Molarity of formic acid = 0.005 moles/1L = 0.005M Molarity of formate = 0.0070moles/1L = 0.007M Now calulate the pKa pKa = - log (Ka) pKa = - log (1.8*10^-4) = 3.74 Use Henderson Hasselbach equation to love for pH pH = pKa + log (Molarity of formate/Molarity of formic acid) pH = 3.74 + log(0.007/0.005) pH = 3.74 + log (1.4) pH = 3.74 + 0.146 pH = 3.89 (b) If the solution was diluted 10000 times, the ratio of molarity of formate to molarity of formic acid will stay the same and the resultant pH would still be 3.89.

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