You are asked to prepare a pH = 4.00 buffer starting from 1.50 L of 0.0200 M solution of benzoic acid (C6H5COOH) and some added sodium benzoate salt (C6H5COONa).
a) Find the pH of the benzoic acid solution BEFORE adding the salt.
b) Use the Henderson-Hasselbalck equation to derive how many moles of the salt you should add.
c) How many grams of the salt should you add?
a) The pH of the benzoic acid solution before adding the salt can be calculated using the Henderson-Hasselbalch equation. The pKa of benzoic acid is 4.20, so the pH of the solution is 4.20.
b) The Henderson-Hasselbalch equation is pH = pKa + log([salt]/[acid]). To calculate the amount of salt needed, we can rearrange the equation to [salt]/[acid] = 10^(pH-pKa). Plugging in the values, we get [salt]/[acid] = 10^(4.00-4.20) = 0.7943. Since the initial concentration of benzoic acid is 0.0200 M, the amount of sodium benzoate needed is 0.0200 x 0.7943 = 0.0159 M.
c) To calculate the amount of sodium benzoate in grams, we can use the molar mass of sodium benzoate, which is 144.11 g/mol. The amount of sodium benzoate needed is 0.0159 mol, so the amount of sodium benzoate in grams is 0.0159 x 144.11 = 2.27 g.
a) To find the pH of the benzoic acid solution, we can use the pKa value of benzoic acid, which is 4.20.
pH = pKa + log ([A-]/[HA])
Since we want to find the pH before adding the salt, we can consider only the benzoic acid.
pH = 4.20 + log ([A-]/[HA])
Since the benzoic acid is a weak acid, it dissociates into its conjugate base, benzoate ion (A-), and a hydronium ion (H+).
Therefore, [A-]/[HA] = [H+]
Since C6H5COOH is initially at a concentration of 0.0200 M, [HA] = 0.0200 M.
Hence, [H+] = 0.0200 M.
pH = -log[H+] = -log(0.0200) = 1.70
Therefore, the pH of the benzoic acid solution before adding the salt is 1.70.
b) The Henderson-Hasselbalck equation can be expressed as:
pH = pKa + log ([A-]/[HA])
Rearranging the equation gives:
[A-]/[HA] = 10^(pH - pKa)
Substituting the given pH of 4.00 and pKa of 4.20 into the equation:
[A-]/[HA] = 10^(4.00 - 4.20) = 10^(-0.20)
[A-]/[HA] = 0.63
This means that for every 0.63 moles of the salt, there is 1 mole of benzoic acid.
Therefore, to achieve the desired pH, we need to add 0.63 moles of the salt (C6H5COONa).
c) The molar mass of C6H5COONa is calculated as follows:
C6H5COONa: C (12.01 g/mol) x 6 + H (1.01 g/mol) x 5 + C (16.00 g/mol) + Na (22.99 g/mol) = 144.11 g/mol
To find the mass of the salt, we multiply the number of moles of the salt by its molar mass:
Mass = moles x molar mass
Mass = 0.63 moles x 144.11 g/mol = 90.8293 g
Therefore, you should add approximately 90.83 grams of the salt (C6H5COONa) to prepare the pH = 4.00 buffer.
a) To find the pH of the benzoic acid solution before adding the salt, we can use the equation for calculating the pH of a weak acid solution. Benzoic acid (C6H5COOH) is a weak acid, so we can assume that it does not fully dissociate in water.
The dissociation reaction of benzoic acid in water can be represented as:
C6H5COOH ⇌ C6H5COO- + H+
The Ka expression for benzoic acid can be written as:
Ka = [C6H5COO-][H+] / [C6H5COOH]
Since the sodium benzoate salt has not been added yet, we don't have any C6H5COO- ions in solution. Therefore, the initial concentration of C6H5COO- is 0.
The initial concentration of C6H5COOH is given as 0.0200 M.
To calculate the initial concentration of H+, we need to assume that benzoic acid is a monoprotic acid. This means that the concentration of H+ ions is equal to the concentration of C6H5COOH.
Therefore, the concentration of H+ ions is 0.0200 M.
To find the pH, we can use the equation:
pH = -log[H+]
Substituting the concentration of H+ ions into the equation, we have:
pH = -log(0.0200) = 1.70
Therefore, the pH of the benzoic acid solution before adding the salt is approximately 1.70.
b) To calculate how many moles of sodium benzoate salt (C6H5COONa) should be added, we can use the Henderson-Hasselbalch equation:
pH = pKa + log([salt]/[acid])
In this case, since we want a pH of 4.00, we need to find the pKa value of benzoic acid.
The pKa value can be found from a table or using the equation:
pKa = -log(Ka)
The Ka value can be obtained from literature sources or experimental data.
Assuming a pKa value of 4.20 for benzoic acid, we can substitute the values into the Henderson-Hasselbalch equation:
4.00 = 4.20 + log([salt]/[acid])
Rearranging the equation, we have:
log([salt]/[acid]) = 4.00 - 4.20 = -0.20
To solve for [salt]/[acid], we can take the antilog of both sides:
[salt]/[acid] = antilog(-0.20)
[salt]/[acid] = 0.63
Therefore, the ratio of [salt] to [acid] should be approximately 0.63.
c) To calculate how many grams of sodium benzoate salt (C6H5COONa) should be added, we need to know the molar mass of the salt.
The molar mass of sodium benzoate can be calculated by adding the molar masses of the individual elements:
C: 12.01 g/mol
H: 1.01 g/mol
Na: 22.99 g/mol
O: 16.00 g/mol
Molar Mass of C6H5COONa = (6 × 12.01) + (5 × 1.01) + 22.99 + 16.00 = 144.10 g/mol
Since the ratio of [salt] to [acid] is 0.63, and the initial concentration of benzoic acid was given as 0.0200 M, we can calculate the concentration of sodium benzoate salt:
[salt] = (0.63) × (0.0200 M) = 0.0126 M
To find the number of moles of sodium benzoate salt, we can multiply the concentration by the volume of the solution (1.50 L):
moles of salt = (0.0126 mol/L) × (1.50 L) = 0.0189 mol
Finally, we can calculate the mass of sodium benzoate salt using the molar mass:
mass of salt = moles of salt × molar mass = 0.0189 mol × 144.10 g/mol = 2.72 g
Therefore, approximately 2.72 grams of sodium benzoate salt should be added.