Question

Formic acid (Ka= 1.7 x 10^-4) and sodium formate were used to prepare a buffer solution...

Formic acid (Ka= 1.7 x 10^-4) and sodium formate were used to prepare a buffer solution of pH 4.31. The concentration of the undissociated formic acid in this buffer solution was 0.80 mol/L. Under these conditions, the molar concentration of formate ion was this: please show work, answer is 1.4 mol/L

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Homework Answers

Answer #1

We will use Henersen Hassel balch equation for buffer solution

According to it the pH of buffer solution is

pH = pKa + log[salt] / [Acid]

pH = 4.31

pKa = -logKa = 3.77

[Salt] = ?

[Acid] = 0.8

on putting values

4.31 = 3.77 + log[salt ] / [acid]

4.31 - 3.77 = log [salt] / [acid]

Taking antilog on both side

3.47 = [salt] /[acid]

[Salt] = 3.47 X [Acid]

[salt] = 3.47 X 0.8 = 2.8

This will be the concentration of salt , Not 1.4 (which is just half of actual value)

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