A buffer is prepared by adding 300.0mL of 2.0M NaOH to 500.0mL of 2.0M CH3COOH. What is the pH of this buffer? ka=1.8x10-5

Amanda is wrong. It is 4.92.

Do the ICE tablee (kinda like DrBob's) then plug it into pH=pKa+log(600/400)
pka=-log(ka)=4.74472

pH=4.92

This one is a little different than your posts under umpteen other screen names. Here you are MAKING the buffer by neutralizing an excess of the acid.

First the mols are:
millimols NasOH = mL x M = 300 x 2.0 = 600
mmols CH3COOH = 500 x 2.0M = 1000
Now make an ICE chart.
.......CH3COOH + NaOH ==> CH3COONa + H2O
initial..1000......0........0........0
add..............600................
change...-600....-600.....+600.....+600
equil....400......0........600......600

Now substitute the millimoles in the ICE chart (the equilibrium line) into the HH equation and solve for pH.
Note: Technically, base/acid in the HH equation goes in in molarity (millimoles/mL)and one should convert 400 mmoles CH3COOH and 600 mmoles CH3COONa to M by 400/800 and 600/800 respectively; however, since the volume of 800 mL is the same for both, the volume cancels, and one can take a short cut and use mmoles alone. Chemically it isn't exactly right but mathematically it is.
Post your work if you get stuck.

Well, first things first, let's gather the necessary information. The initial concentrations of both components in the buffer are 2.0M. Now, let's calculate the moles of NaOH and CH3COOH.

Moles of NaOH = (0.300 L) x (2.0 mol/L) = 0.600 mol
Moles of CH3COOH = (0.500 L) x (2.0 mol/L) = 1.000 mol

Since NaOH is a strong base, it'll be completely ionized in solution, which means it'll produce 0.600 mol of OH-. On the other hand, CH3COOH is a weak acid that doesn't fully dissociate, so it'll produce less than 1.000 mol of H+.

Now, let's calculate the concentration of each component in the final buffer solution:

[OH-] = (0.600 mol) / (0.300 L + 0.500 L) = 0.857 M
[H+] = Since CH3COOH acts as an acid, it'll donate H+ ions. We need to calculate the concentration using the Ka expression.

Ka = [H+][CH3COO-] / [CH3COOH]

Taking the square root of both sides and solving for [H+], we get:

[H+] = √(Ka * [CH3COOH]) = √((1.8x10-5) * 2.0) = √3.6x10-5 = 6x10-3 M

Now, let's calculate the pOH and pH of the buffer:

pOH = -log[OH-] = -log(0.857) = 0.067
pH = 14 - pOH = 14 - 0.067 = 13.933

So, the pH of this buffer solution is approximately 13.933.

To find the pH of a buffer solution, we need to consider the equilibrium between the weak acid (CH3COOH) and its conjugate base (CH3COO-) after they react with the strong base (NaOH). This is an example of an acid-base reaction.

First, let's determine the moles of acetic acid (CH3COOH) and sodium hydroxide (NaOH) in the solution:

Moles of CH3COOH = volume (L) × molarity
= (500.0 mL ÷ 1000 mL/L) × 2.0 M
= 1.0 mol

Moles of NaOH = volume (L) × molarity
= (300.0 mL ÷ 1000 mL/L) × 2.0 M
= 0.6 mol

Since NaOH is a strong base and completely dissociates, it will react fully with acetic acid (CH3COOH) to form water and the acetate ion (CH3COO-). The moles of acetic acid that react with the moles of sodium hydroxide will determine the concentration of CH3COOH and CH3COO- in the buffer solution.

After neutralization, the remaining acetic acid (CH3COOH) concentration in the buffer solution is:

Moles of remaining CH3COOH = initial moles - moles reacted
= 1.0 mol - 0.6 mol
= 0.4 mol

The concentration of CH3COOH can be calculated using the Henderson-Hasselbalch equation:

pH = pKa + log([A-]/[HA])

Where:
pH - the pH of the buffer
pKa - the logarithmic value of the acid dissociation constant (Ka)
[A-] - the concentration of the acetate ion (CH3COO-)
[HA] - the concentration of acetic acid (CH3COOH)

Given that the Ka value for acetic acid (CH3COOH) is 1.8 × 10^-5, we can convert it to pKa:

pKa = -log(Ka)
pKa = -log(1.8 × 10^-5)
pKa = 4.74

Substituting the known values into the Henderson-Hasselbalch equation:

pH = 4.74 + log([CH3COO-]/[CH3COOH])

The initial concentrations of CH3COO- and CH3COOH are equal to the number of moles divided by the total volume (in liters) of the buffer solution:

[CH3COO-] = moles of CH3COO- ÷ total volume (L)
= 0.6 mol ÷ (500 mL ÷ 1000 mL/L + 300 mL ÷ 1000 mL/L)
= 0.6 mol ÷ 0.8 L
= 0.75 M

[CH3COOH] = moles of CH3COOH ÷ total volume (L)
= 0.4 mol ÷ (500 mL ÷ 1000 mL/L + 300 mL ÷ 1000 mL/L)
= 0.4 mol ÷ 0.8 L
= 0.50 M

Now we can substitute these values into the Henderson-Hasselbalch equation:

pH = 4.74 + log(0.75/0.50)
pH = 4.74 + log(1.5)
pH ≈ 4.74 + 0.18
pH ≈ 4.92

Therefore, the pH of this buffer solution is approximately 4.92.

4.57