Mercuric iodate `[Hg_5(IO_6)_2]` reacts with a mixutre of KI and HCl according to the following equation.
`Hg_5(IO_6)_2+34KI+24"HCl"to5K_2HgI_4+8I_2+24KCI+12H_2O`
or `Hg_5(IO_6)_2+34I^(ɵ)+24H^(o+)to5HgI_4^(2-)+8I_2+12H_2O`
The liberated `I_2` is titrated with `Na_2S_2O_3` solution. 1.0 " mL of " `Na_2S_2O_3` is equivalent to 3.992 g of `CuSO_4.5H_2O`. What volume (in mL) of the `Na_2S_2O_3` solution will be required to react with liberated `I_2` from `14.485 g Hg_5(IO_6)_2`.`[Hg=200.5,Cu=63.5,I=127]`?
`Mw(Hg_5(IO_6)_2)=1448.5g`
`Mw(CuSO_4.5H_2O)=249.5g`

Explanation

1 " mol of "`Hg_5(IO_6)_2-=8I_20=16" mol of "CuSO_4.5H_2O`.
(Since 1 " mol of "`I_2` reacts with 2 " mol of "`CuSO_4.5H_2O` in iodometric titration)
`1448.5g of Hg_5(IO_6)_2-=8I_2-=16xx249.5g of CuSO_4.5H_2O`
`3.992 g CuSO_4.5H_2O-=1" mL of " Na_2S_2O_3`
`therefore39.92 g CuSO_4.5H_2O-=10" mL of " Na_2S_2O_3`
`therefore` Volume of `Na_2S_2O_3=10mL`