Qualitative Analysis - Result Question 55
####58. An aqueous solution containing one mole of $HgI _2$ and two moles of $NaI$ is orange in colour. On addition of excess $NaI$ the solution becomes colourless. The orange colour reappears on subsequent addition of $NaOCl$. Explain with equations.
(1999, 3M)
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Solution:
- $NaI$ on reaction with $HgI _2$ gives a complex salt :
$$ 2 NaI+HgI _2 \rightleftharpoons Na _2\left[HgI _4\right] $$
The orange colour is due to residual $HgI _2$. On addition of excess $NaI$, whole $HgI _2$ is converted to complex $Na _2\left[HgI _4\right]$ as colour disappear. The orange colour of $HgI _2$ reappear due to conversion of $Na _2\left[HgI _4\right]$ into $HgI _2$ on treatment with $NaOCl$.
$$ \begin{aligned} 3 Na _2\left[HgI _4\right]+2 NaOCl+2 H _2 O \longrightarrow 3 HgI _2 & +2 NaCl \ & +4 NaOH+2 NaI \end{aligned} $$
