Use the Henderson-Hasselbalch equation.
pH = pKa + log[(base/acid)]
You have solutions of 0.200 M HNO2 and 0.200 M KNO2 (Ka for HNO2 = 4.00 10-4). A buffer of pH 3.000 is needed. What volumes of HNO2 and KNO2 are required to make 1 liter of buffered solution?
4 answers
Howw? How does it apply? I kinda forgot over spring break... How do you find the volume that you need?
in the log part, what dr Bob wrote as log(base/acid), base means concentration of the base, and acid means concentration of acid. concentration is moles/volume. Your two volumes will add to 1 liter.
To start, we assume that the 1 litre of buffer solution requires the mixing of x litres of 0.2M nitrous acid HNO2 and (1 - x) litres of 0.2M potassium nitrite, KNO2. From the Henderson-Hasselbalch equation
pH = pKa + log[NO2-]/[HNO2]
where pKa = - log[4.00 x 10^(-4)] = 3.398 and the required pH = 3.000, so that
[NO2-]/[HNO2] = 10^(-0.398) = 0.400
This equation in itself provides a reasonable estimate of the proportions of HNO2 and KNO2 required, if it is assumed that the dissociation of the acid may be neglected, while that of the salt is complete. (The result is x = 0.714 litres of HNO2.) However, since we know the extent of the dissociation of the nitrous acid, from the pH of the final mixture, we may correct for the first of these assumptions without much difficulty.
From the pH of the buffer, the concentration of hydronium ion is 10^(-3.000) M. In 1 L of buffer, this requires that 10^(-3) mols of HNO3 will dissociate
HNO2 → H+ + NO2-
reducing the amount of HNO2 present from 0.2.x moles to (0.2.x - 0.001). The amount of the conjugate base NO2- will at the same time be increased from [0.2.(1 - x)] moles to [0.2.(1 - x) + 0.001], so that the corrected concentrations for the Henderson-Hasselbalch equations will be
0.400 = [0.2.(1 - x) + 0.001]/[0.2.x - 0.001]
which on multiplying out and rearranging yields x = 0.719 litre of HNO2, with (1 - x) = 0.281 litre of KNO2.
There will of course be some small discrepancy between the calculated pH and that obtained in practice, resulting from assumptions made in the derivation of the Henderson-Hasselbalch equation and uncertainties in the values of associated constants. In addition, nitrous acid and solutions of nitrites are not stable in the medium to long-term.
pH = pKa + log[NO2-]/[HNO2]
where pKa = - log[4.00 x 10^(-4)] = 3.398 and the required pH = 3.000, so that
[NO2-]/[HNO2] = 10^(-0.398) = 0.400
This equation in itself provides a reasonable estimate of the proportions of HNO2 and KNO2 required, if it is assumed that the dissociation of the acid may be neglected, while that of the salt is complete. (The result is x = 0.714 litres of HNO2.) However, since we know the extent of the dissociation of the nitrous acid, from the pH of the final mixture, we may correct for the first of these assumptions without much difficulty.
From the pH of the buffer, the concentration of hydronium ion is 10^(-3.000) M. In 1 L of buffer, this requires that 10^(-3) mols of HNO3 will dissociate
HNO2 → H+ + NO2-
reducing the amount of HNO2 present from 0.2.x moles to (0.2.x - 0.001). The amount of the conjugate base NO2- will at the same time be increased from [0.2.(1 - x)] moles to [0.2.(1 - x) + 0.001], so that the corrected concentrations for the Henderson-Hasselbalch equations will be
0.400 = [0.2.(1 - x) + 0.001]/[0.2.x - 0.001]
which on multiplying out and rearranging yields x = 0.719 litre of HNO2, with (1 - x) = 0.281 litre of KNO2.
There will of course be some small discrepancy between the calculated pH and that obtained in practice, resulting from assumptions made in the derivation of the Henderson-Hasselbalch equation and uncertainties in the values of associated constants. In addition, nitrous acid and solutions of nitrites are not stable in the medium to long-term.
2 answers