Q59 / 180 · JIPMER 2015 · Chemistry

0.001 mole of $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5\left(\mathrm{NO}_3\right)\left(\mathrm{SO}_4\right)\right]$ was passed through a cation exchanger and the acid coming out of it required 20 mL of 0.1 M NaOH for neutralisation. Thus, the complex is

  • Correct answer
    A
    $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5\left(\mathrm{NO}_3\right)\right] \mathrm{SO}_4$
  • B
    $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5\left(\mathrm{SO}_4\right)\right] \mathrm{NO}_3$
  • C
    $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5\right] \mathrm{NO}_3 \cdot \mathrm{SO}_4$
  • D
    None of the above

Solution

Millimoles of acid required for neutralisation $\begin{aligned} & =20 \times 0.1=2 \text { millimoles } \\ & =0.002 \text { moles }\end{aligned}$ If 0.001 mole is neutralising 0.002 moles of base, thus, 0.001 mole of base remains unreacted. Hence, acid coming out should be basic. $\begin{aligned} & {\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5\left(\mathrm{NO}_3\right)\right] \mathrm{SO}_4 \longrightarrow} \\ & {\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5\left(\mathrm{NO}_3\right]^{2+}+\mathrm{SO}_4^{2-}\right.}\end{aligned}$

Hence, the correct answer is option: 'A'.

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