What is the concentration of H+ ions in a buffer solution containing 0.305 M benzoic acid (with ionization constant 6.5 × 10−5)and 0.834 M sodium benzoate (a salt of benzoic acid)?
To calculate the concentration of H+ ions in the buffer solution, we first need to understand the behavior of the benzoic acid and sodium benzoate in water.
Benzoic acid (C6H5COOH) is a weak acid that partially dissociates in water, while sodium benzoate (C6H5COONa) is the salt formed by the reaction between benzoic acid and sodium hydroxide. In water, sodium benzoate completely dissociates into its constituent ions.
The ionization constant (Ka) of benzoic acid is given as 6.5 × 10^−5, which allows us to calculate the concentration of H+ ions using the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
Where:
pH = The pH of the buffer solution
pKa = The negative logarithm of the acid dissociation constant (Ka)
[A-] = Concentration of the conjugate base (sodium benzoate)
[HA] = Concentration of the weak acid (benzoic acid)
Let's calculate the concentration of H+ ions step-by-step:
Step 1: Calculate the pKa from the Ka value:
pKa = -log(Ka) = -log(6.5 × 10^−5)
Step 2: Calculate the concentration ratio [A-]/[HA]:
[A-]/[HA] = [sodium benzoate]/[benzoic acid] = 0.834 M / 0.305 M
Step 3: Calculate the pH:
pH = pKa + log([A-]/[HA])
Now we can proceed with the calculations:
Step 1: Calculate pKa:
pKa = -log(6.5 × 10^−5)
Step 2: Calculate the concentration ratio [A-]/[HA]:
[A-]/[HA] = 0.834 M / 0.305 M
Step 3: Calculate the pH:
pH = pKa + log([A-]/[HA])
Once you provide the Ka value, I can assist you further in calculating the pH.
To find the concentration of H+ ions in the buffer solution, we need to consider the dissociation of benzoic acid and the hydrolysis of sodium benzoate.
1. Dissociation of benzoic acid:
Benzoic acid (C6H5COOH) can partially dissociate in water, releasing H+ ions.
The ionization constant (Ka) for benzoic acid is given as 6.5 × 10−5.
The dissociation equation for benzoic acid is:
C6H5COOH ⇌ C6H5COO- + H+
2. Hydrolysis of sodium benzoate:
Sodium benzoate (C6H5COONa) undergoes hydrolysis in water, producing OH- ions.
The hydrolysis equation for sodium benzoate is:
C6H5COONa + H2O ⇌ C6H5COOH + NaOH
In a buffer solution, both processes occur simultaneously. The benzoate ions (C6H5COO-) resulting from the dissociation of benzoic acid react with the OH- ions produced from the hydrolysis of sodium benzoate, forming water and benzoic acid.
The key concept in buffer solutions is the equilibrium between the weak acid and its conjugate base. In this case, the weak acid is benzoic acid (C6H5COOH), and its conjugate base is the benzoate ion (C6H5COO-).
To calculate the concentration of H+ ions in the buffer solution, we need to consider the equilibrium expression (Ka) for benzoic acid:
Ka = [C6H5COO-][H+] / [C6H5COOH]
We are given the initial concentrations of benzoic acid and sodium benzoate: 0.305 M and 0.834 M, respectively.
Now, let's assume the change in concentration of benzoic acid is −x, and the change in concentration of benzoate ions is +x. This is based on the stoichiometry of the equilibrium equation.
At equilibrium, the concentration of benzoic acid will be (0.305 − x) M, the concentration of benzoate ions will be (0.834 + x) M, and the concentration of H+ ions will be x M.
Substituting these values into the equilibrium expression:
Ka = [(0.834 + x)(x)] / (0.305 - x)
Now, solve this quadratic equation to find the value of x, which corresponds to the concentration of H+ ions in the buffer solution.
Using this value of x, you can find the concentration of H+ ions in the buffer solution.