Answer:
Buffer B has the highest buffer capacity.
Buffer C has the lowest buffer capacity.
Explanation:
An effective weak acid-conjugate base buffer should have pH equal to
of the weak acid. For buffers with the same pH, higher the concentrations of the components in a buffer, higher will the buffer capacity.
Acetic acid is a weak acid and
is the conjugate base So, all the given buffers are weak acid-conjugate base buffers. The pH of these buffers are expressed as (Henderson-Hasselbalch):
![pH=pK_{a}(CH_{3}COOH)+log\frac{[CH_{3}COO^{-}]}{[CH_{3}COOH]}](https://tex.z-dn.net/?f=pH%3DpK_%7Ba%7D%28CH_%7B3%7DCOOH%29%2Blog%5Cfrac%7B%5BCH_%7B3%7DCOO%5E%7B-%7D%5D%7D%7B%5BCH_%7B3%7DCOOH%5D%7D)

Buffer A: 
Buffer B: 
Buffer C: 
So, both buffer A and buffer B has same pH value which is also equal to
. Buffer B has higher concentrations of the components as compared to buffer A, Hence, buffer B has the highest buffer capacity.
The pH of buffer C is far away from
. Therefore, buffer C has the lowest buffer capacity.
Answer:
The final temperature is 39.58 degree Celsius
Explanation:
As we know
Q = m * c * change in temperature
Specific heat of water (c) = 4.2 joules per gram per Celsius degree
Substituting the given values we get -
5750 = 335 * 4.2 * (X - 35.5)
X = 39.58 degree Celsius
A joule is defined as the amount of energy transferred to an object when a force of one newton acts on the object in the direction of its motion through a distance of one meter (1 newton meter or Nm)
The solution would be like this for this specific problem:
Given:
pH of a 0.55 M hypobromous
acid (HBrO) at 25.0 °C = 4.48
[H+] = 10^-4.48 = 3.31 x
10^-5 M = [BrO-] <span>
Ka = (3.31 x 10^-5)^2 / 0.55 = 2 x 10^-9</span>
To add, Hypobromous Acid does not require acid
adjustment, which is necessary for chlorine-based product and is stable and
effective in pH ranges of 5-9.<span>
</span>Hypobromous Acid combines with organic
compounds to form a bromamine. Chlorine also combines with the same organic
compounds to form a chloramine. <span>It is also
one of the least expensive intervention antimicrobial compounds available.</span>