Propionic acid, C2H5COOH, is an oily liquid that has a slightly pungent, disagreeable rancid odor. It is found in dairy products in small amounts. Its sodium salt, sodium propionate, C2H5COONa, has wide application as a food preservative, fungicide and mold inhibitor. What is the ratio of propionic acid to sodium propionate in a buffer solution that has a pH of 4.26, if the Ka of propionic acid is 1.38 x 10-5? If exactly 1.0 liters of the buffer solution is to be prepared, how many grams of C2H5COONa will be required?
A buffer is a solution where both an weak acid and its salt os coonjugat ebase is present. Now in our given buffer, the weak acid is C2H5COOH ans its salt of conjugate base is C2H5COONa
Now the pH of a buffer is calculated by Henderson - Hasselbach equation
pH = pKa + log [salt]/[acid]
where pKa = -log Ka
For our given buffer, pH = pKa + log [C2H5COONa] / [C2H5COOH]
Now putting the given values-
pH = pKa + log [C2H5COONa] / [C2H5COOH]
4.26 = - log Ka + log [C2H5COONa] / [C2H5COOH]
4.26 = - log (1.38 x 10-5) + log [C2H5COONa] / [C2H5COOH]
log [C2H5COONa] / [C2H5COOH] = 4.26 + log (1.38 x 10-5)
log [C2H5COONa] / [C2H5COOH] = 4.26 + (-4.86)
log [C2H5COONa] / [C2H5COOH] = 4.26 - 4.86
log [C2H5COONa] / [C2H5COOH] = -0.6
[C2H5COONa] / [C2H5COOH] = 10-0.6
[C2H5COONa] / [C2H5COOH] = 0.25 = 1/4
So the ratio of propionic acid to sodium propionate = [C2H5COOH] / [C2H5COONa] = 4:1
Now if we take 1L of solution, then concentration of [C2H5COOH] = 4M
So moles of C2H5COOH = concentration * volume
= 4M * 1L
= 4 mols/L * 1L
= 4 mols
So mass of C2H5COOH = moles * molar mass
= 4 moles * 74.08 g/mol
= 296.32 g
So 296.32 grams of C2H5COONa will be required to prepare 1L of buffer
Propionic acid, C2H5COOH, is an oily liquid that has a slightly pungent, disagreeable rancid odor. It...
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