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:
[tex]H2SO3\leftrightarrow H^{+}+HSO3^{-}[/tex] --------(1)
where, Ka1 = 1.5 x 10–2 or pKa1 = 1.824
[tex]HSO3^{-}\leftrightarrow H^{+}+SO3^{2-}[/tex] --------(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
[tex]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[/tex]
Step 2:
For the second equivalence point setup an ICE table:
[tex]HSO3^{-}+OH^{-}\leftrightarrow H2O+SO3^{2-}[/tex]
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:
[tex]pH = pKa2 + log\frac{[SO3]^{2-} }{[HSO3]^{-} } \\6.247 = 7.00 + log\frac{x}{(1.98-x)} \\x=0.634 moles[/tex]
Volume of NaOH required is:
[tex]\frac{0.634\ moles}{0.615 moles/L}=0.389L[/tex]
Step 3:
Total volume of NaOH required = 3.22+0.389 =3.61 L
