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If a buffer solution is 0.100 M in a weak acid ( 𝐾a=4.9×10−5) and 0.440 M in its conjugate base, what is the pH?

I figured this one out for myself haha... directions as follows:

1. -log(Ka)=pKa...-log(4.9x10-5)=4.3098...
2. pH=pKa+log(cA/cHA)
3. pH=4.3098...+log(.440/.100)
ans=4.9532... or 5.0

I didn't do the math but your process is on the mark. Good work.

To find the pH of a buffer solution, we need to make use of the Henderson-Hasselbalch equation:

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

In this equation:
- pH is the negative logarithm of the hydrogen ion concentration.
- pKa is the negative logarithm of the acid dissociation constant.
- [A-] is the concentration of the conjugate base.
- [HA] is the concentration of the weak acid.

Given that the buffer solution is 0.100 M in the weak acid and 0.440 M in its conjugate base, we can substitute these values into the Henderson-Hasselbalch equation:

pH = pKa + log([A-]/[HA])
= -log(4.9×10^(-5)) + log(0.440/0.100)

Now, let's calculate this:

pH ≈ -log(4.9×10^(-5)) + log(4.4)
≈ -(-4.309) + 0.643
≈ 4.309 + 0.643
≈ 4.952

Therefore, the pH of the buffer solution is approximately 4.952.

To find the pH of a buffer solution, you can use the Henderson-Hasselbalch equation, which is given by:

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

Where:
pH is the unknown pH value of the buffer solution.
pKa is the negative logarithm of the acid dissociation constant (Ka) of the weak acid.
[A-] is the concentration of the conjugate base.
[HA] is the concentration of the weak acid.

In this case:
pKa = -log(4.9×10−5) = approximately 4.31 (assuming the Ka value you provided is correct)

[A-] = 0.440 M (concentration of the conjugate base)
[HA] = 0.100 M (concentration of the weak acid)

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

pH = 4.31 + log(0.440/0.100)

Calculating the logarithm:

pH = 4.31 + log(4.4)

Finally, calculate the pH:

pH ≈ 4.31 + 0.643

pH ≈ 4.95

Therefore, the pH of the buffer solution is approximately 4.95.