enzoic acid is a weak monoprotic acid with Ka = 6.5×10-5 M. NaOH(s) was gradually added to 1.00 L of 8.66×10-2 M benzoic acid.
1. Calculate the pH of the solution before the addition of the base.
2. Calculate the pH of the solution after the addition of 5.20×10-2 mol of NaOH(s).
To answer these questions, we need to consider the reaction between benzoic acid (C6H5COOH) and sodium hydroxide (NaOH). Benzoic acid is a weak acid that partially ionizes in water, and sodium hydroxide is a strong base that fully dissociates in water. The balanced chemical equation for the reaction is:
C6H5COOH + NaOH → C6H5COONa + H2O
1. Calculate the pH of the solution before the addition of the base:
Before the addition of NaOH, only benzoic acid is present in the solution. We can assume it to be fully ionized since it is a weak acid.
[H+] = [C6H5COOH] = 8.66×10^-2 M
To calculate the pH, we can use the equation:
pH = -log[H+]
So, the pH of the solution before the addition of the base is:
pH = -log(8.66×10^-2) = 1.063
2. Calculate the pH of the solution after the addition of 5.20×10^-2 mol of NaOH(s):
To calculate the pH after the addition of the base, we need to consider the remaining concentration of benzoic acid and the concentration of hydroxide ions (OH-) from the dissociation of sodium hydroxide.
We start with the remaining concentration of benzoic acid:
[C6H5COOH] = 8.66×10^-2 M - (5.20×10^-2 mol / 1 L) = 8.66×10^-2 M - 5.20×10^-2 M = 3.46×10^-2 M
Next, we calculate the concentration of hydroxide ions (OH-) based on the moles of NaOH added:
[OH-] = (5.20×10^-2 mol / 1 L) = 5.20×10^-2 M
Since benzoic acid is a weak acid, we need to consider the reaction between the hydroxide ions and the weak acid. The benzoic acid will react with the hydroxide ions to form water and benzoate ions (C6H5COO-).
C6H5COOH + OH- → C6H5COO- + H2O
By using the equilibrium expression for this reaction, we can write:
Ka = [C6H5COO-] [H+] / [C6H5COOH]
Since [C6H5COOH] is much greater than [C6H5COO-], we can assume that [C6H5COOH] - [C6H5COO-] ≈ [C6H5COOH]. Thus, we can simplify the equation:
Ka = [C6H5COO-] [H+] / (3.46×10^-2 M)
Now, let's rearrange the equation and solve for [H+]:
[H+] = (Ka * [C6H5COOH]) / [C6H5COO-]
[H+] = (6.5×10^-5 M * 3.46×10^-2 M) / (5.20×10^-2 M)
[H+] ≈ 4.33×10^-5 M
Finally, we can calculate the pH using the equation:
pH = -log[H+]
So, the pH of the solution after the addition of 5.20×10^-2 mol of NaOH(s) is:
pH = -log(4.33×10^-5) ≈ 4.36
What is your problem with this bekah?
For a you must have worked a hundred of weak acids problems with a known starting molarity.
At first glance it looks like 2 is a buffer problem that can be solved with the Henderson-Hasselbalch equation.