Answer : The pH of buffer is 9.06.
Explanation : Given,
Concentration of HBrO = 0.34 M
Concentration of KBrO = 0.89 M
Now we have to calculate the pH of buffer.
Using Henderson Hesselbach equation :
![pH=pK_a+\log \frac{[Salt]}{[Acid]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%20%5Cfrac%7B%5BSalt%5D%7D%7B%5BAcid%5D%7D)
![pH=pK_a+\log \frac{[KBrO]}{[HBrO]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%20%5Cfrac%7B%5BKBrO%5D%7D%7B%5BHBrO%5D%7D)
Now put all the given values in this expression, we get:
Therefore, the pH of buffer is 9.06.
Let us say that R is the major enantiomer, while
S is the minor enantiomer, therefore the formula for enantiomeric excess (ee)
is:
ee = (R – S) * 100%
Let us further say that the fraction of R is x (R
= x), and therefore fraction of S is 1 – x (S = 1 – x), therefore:
75 = (x – (1 – x)) * 100
75 = 100 x – 100 + 100 x
200 x = 175
x = 0.875
Summary of answers:
R = major enantiomer = 0.875 or 87.5%
<span>S = minor enantiomer = (1 – 0.875) = 0.125 or
12.5%</span>
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