You prepare a buffer solution by dissolving 2.00 g each of benzoic acid, C6H5COOH and sodium benzoate, NaC6H5COO in 750.0 mL of water. What is the pH of this buffer? Assume that the solution's volume is 750.0 mL.
To determine the pH of a buffer solution, you need to consider the acid-base equilibrium that exists between the weak acid (benzoic acid, C6H5COOH) and its conjugate base (sodium benzoate, NaC6H5COO).
The pKa value of benzoic acid is required to calculate the buffer pH. In this case, the pKa value is not given, so we need to look it up in a reliable source, such as a textbook or chemical database. The pKa of benzoic acid is approximately 4.19.
Next, we need to determine the ratio of the concentration of the weak acid to its conjugate base in the buffer solution. We can calculate this using the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
Where [A-] represents the concentration of the conjugate base (sodium benzoate) and [HA] represents the concentration of the weak acid (benzoic acid).
Given that we dissolved 2.00 g of both benzoic acid and sodium benzoate in 750.0 mL of water, we need to convert these amounts into concentrations (in moles per liter) by dividing by their respective molar masses.
- Molar mass of benzoic acid (C6H5COOH) = 122.12 g/mol
- Molar mass of sodium benzoate (NaC6H5COO) = 144.11 g/mol
Converting the mass of benzoic acid to moles:
Number of moles of benzoic acid = (mass of benzoic acid) / (molar mass of benzoic acid)
= 2.00 g / 122.12 g/mol
= 0.01639 mol
Converting the mass of sodium benzoate to moles:
Number of moles of sodium benzoate = (mass of sodium benzoate) / (molar mass of sodium benzoate)
= 2.00 g / 144.11 g/mol
= 0.01387 mol
Now, we have the concentrations of both [A-] and [HA] in the buffer solution:
[A-] = (moles of sodium benzoate) / (volume of solution in liters)
= 0.01387 mol / 0.75 L
= 0.0185 M
[HA] = (moles of benzoic acid) / (volume of solution in liters)
= 0.01639 mol / 0.75 L
= 0.0219 M
Substituting these values into the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
= 4.19 + log(0.0185/0.0219)
≈ 4.19 + log(0.8436)
≈ 4.19 - 0.0746
≈ 4.12
Therefore, the pH of this buffer solution is approximately 4.12.
mols = grams/molar mass
and M = mols/L
Convert 2.00 g each into mols with the above formula, then use the second one to find M.
Plug these into the Henderson-Hasselbalch equation and solve for pH. Post your work if you get stuck.