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
Answer:
Atom
Explanation:
atom: The basic unit of matter; the smallest unit of an element, having all the characteristics of that element; consists of negatively-charged electrons and a positively-charged center called a nucleus.
That is, by definition, true.
Answer: The last electron will be filled in first orbital of 3p sub-shell.
Explanation: Filling of electrons in orbitals is done by using Hund's Rule.
Hund's rule states that the electron will be singly occupied in the orbital of the sub-shell before any orbital is doubly occupied.
For filling up of the electrons in Sulfur atom having 16 electrons. First 10 electrons will completely fill according to Aufbau's Rule in 1s, 2s and 2p sub-shells and last 6 electrons are the valence electrons which will be filled in the order of 3s and then 3p.
3s sub-shell will be fully filled and the orbitals of 3p sub-shell will be first singly occupied and then pairing will take place. Hence, the last electron will be filled in the first orbital of 3p-sub-shell.