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
![i.e. \ 5.13\ moles/L*0.385L = moles\ NaOH\\therefore, \ moles\ NaOH = 1.98\ moles\\V(NaOH)\ required = \frac{1.98\ moles}{0.615\ moles/L} =3.22L](https://tex.z-dn.net/?f=i.e.%20%5C%205.13%5C%20%20moles%2FL%2A0.385L%20%3D%20moles%5C%20%20NaOH%5C%5Ctherefore%2C%20%5C%20moles%5C%20%20NaOH%20%3D%201.98%5C%20moles%5C%5CV%28NaOH%29%5C%20required%20%3D%20%5Cfrac%7B1.98%5C%20moles%7D%7B0.615%5C%20moles%2FL%7D%20%3D3.22L)
Step 2:
For the second equivalence point setup an ICE table:
![HSO3^{-}+OH^{-}\leftrightarrow H2O+SO3^{2-}](https://tex.z-dn.net/?f=HSO3%5E%7B-%7D%2BOH%5E%7B-%7D%5Cleftrightarrow%20H2O%2BSO3%5E%7B2-%7D)
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:
![\frac{0.634\ moles}{0.615 moles/L}=0.389L](https://tex.z-dn.net/?f=%5Cfrac%7B0.634%5C%20moles%7D%7B0.615%20moles%2FL%7D%3D0.389L)
Step 3:
Total volume of NaOH required = 3.22+0.389 =3.61 L