Answer:
Volume of NaOH required = 3.61 L
Explanation:
H2SO3 is a diprotic acid i.e. it will have two dissociation constants given as follows:
--------(1)
where, Ka1 = 1.5 x 10–2 or pKa1 = 1.824
--------(2)
where, Ka2 = 1.0 x 10–7 or pKa2 = 7.000
The required pH = 6.247 which is beyond the first equivalence point but within the second equivalence point.
Step 1:
Based on equation(1), at the first eq point:
moles of H2SO3 = moles of NaOH

Step 2:
For the second equivalence point setup an ICE table:

Initial 1.98 ? 0
Change -x -x x
Equil 1.98-x ?-x x
Here, ?-x =0 i.e. amount of OH- = x
Based on the Henderson buffer equation:
![pH = pKa2 + log\frac{[SO3]^{2-} }{[HSO3]^{-} } \\6.247 = 7.00 + log\frac{x}{(1.98-x)} \\x=0.634 moles](https://tex.z-dn.net/?f=pH%20%3D%20pKa2%20%2B%20log%5Cfrac%7B%5BSO3%5D%5E%7B2-%7D%20%7D%7B%5BHSO3%5D%5E%7B-%7D%20%7D%20%5C%5C6.247%20%3D%207.00%20%2B%20log%5Cfrac%7Bx%7D%7B%281.98-x%29%7D%20%5C%5Cx%3D0.634%20moles)
Volume of NaOH required is:

Step 3:
Total volume of NaOH required = 3.22+0.389 =3.61 L
Explanation:
Let relative ratio of one isotope (62.94 u) be X
Then, relative ratio of other isotope (64.93) will be (1 - X)
Now,
(62.94)x + (64.93)(1 - x) = 63.55
1.99x = 1.38
X = 0.69
<u>Relative Abundance</u> :
(62.94 u) isotope = 69 %
(64.93 u) isotope = 31 %