You are carrying out the titration of 100.0 mL of 1.000M acetic acid with 1.000 M sodium hydroxide. Ka = 1.76x10^-5 of acetic acid.
(a) Calculate the initial pH of your acetic acid sample.
(b) Calculate the pH of the solution after the addition of 25.0mL of NaOH.
For (a) should I just take the -log of the given Ka value, which is 4.75?
or should I do this:
CH3COOH <--> CH3COO- + H+
1.0 0 , 0
-x +x , +x
1.0-x x , x
1.0-x can be approximated to 1.0
so, x^2/1.0=1.76x10^-5
x = 4.20x10^-3, and x = [H+]
-log [H+] = pH = 2.38
I'm not sure if should just be 2.38 or 4.75
then part B, I set it up like this:
so convert both CH3COOH and NaOH to moles
.1 CH3COOH and .025 NaOH
CH3COOH + NaOH <--> CH3COONa + H2O
I subtracted moles of NaOH from CH3COOH on the reactants side, and added moles of NaOH to CH3COONa (which starts at zero) on the products side
.1-0.025 = 0.075 of CH3COOH, and 0+0.025 = 0.025 of CH3COO-
because these are conjugates, I can use Henderson-Hasselbalch:
pH = pKa + log 0.025/0.075
pH = 4.27
Is this correct? Can the pH go down when I'm adding a strong base (even if it's a smaller amount) to a weak acid?
For a, the latter procedure is correct.
For b, I came out with 4.277 which I would round to 4.28 but check my arithmetic. Yes, the pH can do down because you are neutralizing 1/4 of the acid and the extra conjugate base that's formed suppresses the ionization of acetic acid more, also. Good work.
Donna, I put my thinking cap on wrong this morning. There's nothing wrong with your answers. The latter method is the one to use for the first part and your second part is done very well, although, I arrived at 4.28 instead of 4.27. My error is in the explanation of acetic acid becoming more acid. It doesn't. The pH of the acid, alone, is 2.38 in the first part and becomes 4.28 when NaOH is added to form the sodium acetate salt. The acetic acid concentration is less and the pH is greater. It is more basic in the buffer solution than in 1 M acetic acid. Sorry about that. I now see you were comparing 4.75 with 4.28 but the 4.75 was never there since that isn't the way to approach the problem.
(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.