50 mL of 0.1 M acetic acid is mixed with 50 mL of 0.1 M sodium acetate (the conjugate base). The Ka of acetic acid is approximat
ely 1. 74 X 10 -5. What is the pH of the resulting solution?
1 answer:
Answer:
4.76
Explanation:
In this case, we have to start with the <u>buffer system</u>:

We have an acid (
) and a base (
). Therefore we can write the <u>henderson-hasselbach reaction</u>:
![pH~=~pKa+Log\frac{[CH_3COO^-]}{[CH_3COOH]}](https://tex.z-dn.net/?f=pH~%3D~pKa%2BLog%5Cfrac%7B%5BCH_3COO%5E-%5D%7D%7B%5BCH_3COOH%5D%7D)
If we want to calculate the pH, we have to <u>calculate the pKa</u>:

According to the problem, we have the <u>same concentration</u> for the acid and the base 0.1M. Therefore:
![[CH_3COO^-]=[CH_3COOH]](https://tex.z-dn.net/?f=%5BCH_3COO%5E-%5D%3D%5BCH_3COOH%5D)
If we divide:
![\frac{[CH_3COO^-]}{[CH_3COOH]}~=~1](https://tex.z-dn.net/?f=%5Cfrac%7B%5BCH_3COO%5E-%5D%7D%7B%5BCH_3COOH%5D%7D~%3D~1)
If we do the Log of 1:

So:

With this in mind, the pH is 4.76.
I hope it helps!
You might be interested in
Answer: B, E and F.
Explanation:
Just took a test on it and those were the correct answers.
I believe it would be Boyle's Law
(P [is proportional to] 1/V)
P being pressure
V being volume
Answer:
9
option d 9
pH of a 10^-5 M HCI solution is 9
Protons=7
electrons=7
nuetrons=7
Answer:
8 Nitrogen atoms
Explanation:
Step 1: Define
Chemical Formula - C₈H₁₀(N₄O₂)₂
Step 2: Multiply
(N₄)₂ = N₈