(a) To calculate the initial pH of the acetic acid sample, you need to use the dissociation constant (Ka) of acetic acid. The equation for the dissociation of acetic acid is:
CH3COOH ⇌ CH3COO- + H+
The initial concentration of acetic acid is given as 1.000 M. Let's assume the concentration of the dissociated acetic acid (CH3COO-) and the concentration of the hydrogen ion (H+) are both x.
Using the equation for the dissociation of acetic acid, you can set up the following expressions for the concentrations:
[CH3COOH] = 1.000 - x
[CH3COO-] = x
[H+] = x
The Ka expression for acetic acid is given as 1.76x10^-5, which is equal to [CH3COO-][H+]/[CH3COOH]. Substituting the expressions for the concentrations, you can solve for x:
1.76x10^-5 = (x)(x)/(1.000 - x)
Note that the value of (1.000 - x) can be approximated to 1.000 since x is expected to be small during the initial stage of acid dissociation.
Simplifying the equation gives: x^2/1.000 ≈ 1.76x10^-5
Solving for x, you will find that x is approximately 4.20x10^-3 M.
Since x represents the concentration of H+, you can calculate the pH as -log[H+], which gives a value of 2.38.
Therefore, the correct answer for part (a) is pH = 2.38.
(b) For part (b), you are adding 25.0 mL of 1.000 M NaOH to the acetic acid solution. First, convert the volume of NaOH solution to moles:
(25.0 mL)(1.000 M) = 0.025 moles of NaOH
In the balanced equation for the reaction between acetic acid and NaOH:
CH3COOH + NaOH ⇌ CH3COONa + H2O
the coefficient ratio between acetic acid and NaOH is 1:1. Therefore, the moles of acetic acid (CH3COOH) remaining after the reaction will be the initial moles of acetic acid (0.100 moles) minus the moles of NaOH added (0.025 moles):
0.100 - 0.025 = 0.075 moles of CH3COOH
To calculate the concentration of CH3COO-, divide the moles of CH3COO- by the total volume of the solution in liters:
0.025 moles / (0.100 L + 0.025 L) = 0.025 moles / 0.125 L = 0.200 M
Now you can use the Henderson-Hasselbalch equation to calculate the pH:
pH = pKa + log([CH3COO-]/[CH3COOH])
Since the Ka of acetic acid (pKa) was not given directly, you can use the Ka value from part (a) to calculate pKa:
pKa = -log(Ka) = -log(1.76x10^-5) ≈ 4.75
Plugging in the values:
pH = 4.75 + log(0.200/0.075) = 4.75 + log(2.67) ≈ 4.75 + 0.426 = 5.18
Therefore, the correct answer for part (b) is pH = 5.18.
Yes, the pH can go down when you are adding a strong base to a weak acid. In this case, the strong base (NaOH) reacts with the weak acid (acetic acid), reducing the concentration of acetic acid and increasing the concentration of its conjugate base (CH3COO-). As a result, the pH increases towards the basic side.