Answer:

Explanation:
Hello!
In this case, we can divide the problem in two steps:
1. Dilution to 278 mL: here, the initial concentration and volume are 1.20 M and 52.0 mL respectively, and a final volume of 278 mL, it means that the moles remain the same so we can write:

So we solve for C2:

2. Now, since 111 mL of water is added, we compute the final volume, V3:

So, the final concentration of the 139 mL portion is:

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<u>Answer:</u> The pH of the buffer is 5.25
<u>Explanation:</u>
Let the volume of buffer solution be V
We know that:

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:
![pH=pK_a+\log(\frac{[\text{conjugate base}]}{[acid]})](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%28%5Cfrac%7B%5B%5Ctext%7Bconjugate%20base%7D%5D%7D%7B%5Bacid%5D%7D%29)
We are given:
= negative logarithm of acid dissociation constant of weak acid = 4.90
![[\text{conjugate base}]=\frac{2.25}{V}](https://tex.z-dn.net/?f=%5B%5Ctext%7Bconjugate%20base%7D%5D%3D%5Cfrac%7B2.25%7D%7BV%7D)
![[acid]=\frac{1.00}{V}](https://tex.z-dn.net/?f=%5Bacid%5D%3D%5Cfrac%7B1.00%7D%7BV%7D)
pH = ?
Putting values in above equation, we get:

Hence, the pH of the buffer is 5.25
Answer:
Explanation:
The usefulness of a buffer is its ability to resist changes in pH when small quantities of base or acid are added to it. This ability is the consequence of having both the conjugate base and the weak acid present in solution which will consume the added base or acid.
This capacity is lost if the ratio of the concentration of conjugate base to the concentration of weak acid differ by an order of magnitude. Since buffers having ratios differing by more will have their pH driven by either the weak acid or its conjugate base .
From the Henderson-Hasselbach equation we have that
pH = pKa + log [A⁻]/[HA]
thus
0.1 ≤ [A⁻]/[HA] ≤ 10
Therefore the log of this range is -1 to 1, and the pH will have a useful range of within +/- 1 the pKa of the buffer.
Now we are equipped to answer our question:
pH range = 3.9 +/- 1 = 2.9 through 4.9