You are asked to prepare a pH=4.00 buffer starting from 1.50 L of 0.0200 \; M solution of benzoic acid C_6 H_5COOH and an excess of sodium benzoate C_ H_5 COONa.
How many grams of sodium benzoate should be added to prepare the buffer? Neglect the small volume change that occurs when the sodium benzoate is added. (in grams)
Use the HH equation to determine mols sodium benzoate, then mols = grams/molar mass.
To prepare a buffer solution with a desired pH, we can use the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
Where:
pH = desired pH
pKa = acid dissociation constant of the weak acid
[A-] = concentration of the conjugate base
[HA] = concentration of the weak acid
In this case, the weak acid is benzoic acid (C6H5COOH) and the conjugate base is sodium benzoate (C6H5COONa).
To calculate the grams of sodium benzoate (C6H5COONa) needed, we need to do the following steps:
Step 1: Calculate the moles of benzoic acid (C6H5COOH)
Given:
Volume of benzoic acid solution (V) = 1.50 L
Concentration of benzoic acid (C6H5COOH) = 0.0200 M
Moles of benzoic acid (C6H5COOH) = Volume (V) x Concentration (C)
Moles of benzoic acid (C6H5COOH) = 1.50 L x 0.0200 M
Step 2: Calculate the grams of benzoic acid (C6H5COOH)
To calculate the grams of benzoic acid, we need the molar mass of benzoic acid.
Molar mass of benzoic acid (C6H5COOH) = (6 x Atomic mass of carbon) + (5 x Atomic mass of hydrogen) + (2 x Atomic mass of oxygen)
Atomic mass of carbon (C) = 12.01 g/mol
Atomic mass of hydrogen (H) = 1.008 g/mol
Atomic mass of oxygen (O) = 16.00 g/mol
Molar mass of benzoic acid (C6H5COOH) = (6 x 12.01 g/mol) + (5 x 1.008 g/mol) + (2 x 16.00 g/mol)
Using the molar mass, we can now calculate the grams of benzoic acid:
Grams of benzoic acid (C6H5COOH) = Moles of benzoic acid (C6H5COOH) x Molar mass of benzoic acid (C6H5COOH)
Step 3: Calculate the grams of sodium benzoate (C6H5COONa)
Since benzoic acid (C6H5COOH) and sodium benzoate (C6H5COONa) are in a 1:1 ratio, the grams of sodium benzoate needed will be the same as the grams of benzoic acid (C6H5COOH).
Therefore, Grams of sodium benzoate (C6H5COONa) = Grams of benzoic acid (C6H5COOH).
You can now perform the calculations using the given values.
To determine the number of grams of sodium benzoate that should be added to prepare the buffer, we need to consider the balanced chemical equation for the dissociation of benzoic acid:
C6H5COOH ⇌ C6H5COO- + H+
In this equation, benzoic acid (C6H5COOH) dissociates into the benzoate ion (C6H5COO-) and a hydrogen ion (H+).
A buffer solution is formed by a weak acid and its conjugate base or a weak base and its conjugate acid. In this case, benzoic acid (C6H5COOH) is the weak acid and sodium benzoate (C6H5COONa) is the conjugate base.
The Henderson-Hasselbalch equation relates the pH of a buffer solution to the concentrations of the weak acid and its conjugate base:
pH = pKa + log([conjugate base]/[weak acid])
Here, pKa is the logarithmic acid dissociation constant, which is a measure of the strength of the acid. The pKa value for benzoic acid is given as 4.20 in many sources.
We are given that the desired pH of the buffer solution is 4.00. Now, we can determine the ratio of [conjugate base]/[weak acid] using the Henderson-Hasselbalch equation:
4.00 = 4.20 + log([conjugate base]/[weak acid])
Rearranging the equation:
log([conjugate base]/[weak acid]) = 4.00 - 4.20
log([conjugate base]/[weak acid]) = -0.20
Using the properties of logarithms, we can rewrite the equation as an exponential equation:
[conjugate base]/[weak acid] = 10^(-0.20)
[conjugate base]/[weak acid] = 0.100
In a solution with a concentration of weak acid equal to 0.0200 M, the concentration of the conjugate base will be (0.100)(0.0200) = 0.00200 M.
We are given that we have 1.50 L of the 0.0200 M benzoic acid solution. Therefore, the number of moles of benzoate that would be in this volume is:
moles of benzoate = (0.00200 M)(1.50 L) = 0.00300 moles
To determine the mass of sodium benzoate (C6H5COONa) required, we need to convert the moles of benzoate to grams. The molar mass of sodium benzoate is:
[molar mass of sodium (Na)] + [molar mass of benzoate (C6H5COO-)] = (22.99 g/mol) + (121.11 g/mol) = 144.10 g/mol
Now, we can calculate the mass of sodium benzoate:
mass of sodium benzoate = (0.00300 moles)(144.10 g/mol) = 0.432 g
Therefore, approximately 0.432 grams of sodium benzoate should be added to prepare the pH = 4.00 buffer solution.