Which of the following will be the most effective in the coagulation of ${\rm{Fe}}{\left( {{\rm{OH}}} \right)_3}$ sol?

${\rm{M}}{{\rm{g}}_3}{\left( {{\rm{P}}{{\rm{O}}_4}} \right)_2}$

NEET UG CHEMISTRY MCQ: SURFACE CHEMISTRY

1 Easy

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Application +4 | -1

Which of the following will be the most effective in the coagulation of

\mathrm{Fe}{\left( {\mathrm{OH}} \right)_3}

sol?

View answer and explanation

Correct option: D

\mathrm{M}{\mathrm{g}_3}{\left( {\mathrm{P}{\mathrm{O}_4}} \right)_2}

Option D (

\mathrm{M}{\mathrm{g}_3}{\left( {\mathrm{P}{\mathrm{O}_4}} \right)_2}

) is correct because, according to the Hardy-Schulze rule, the ion with the highest charge will have the maximum coagulation power. Since

\mathrm{Fe}{\left( {\mathrm{OH}} \right)_3}

is a positive sol, we need to look for the anion with the highest charge, which is

PO_4^{3-}

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10^{-3} \mathrm{M}

1 \mathrm{~mm}^3

2 \times 10^{13}

\mathrm{N}_{\mathrm{A}}=6 \times 10^{23}

\log \frac{{x}}{{m}}

\log {P}

\theta=45^{\circ}

0.5

\log {k}= 0.699,

\mathrm{g}

\log \;x/m\mathrm{ } \;vs.

\log \;P

19\;\mathrm{g/c}{\mathrm{m}^3}

{1.9 \times {{10}^{ - 4}}\;\mathrm{g}}

10\;\mathrm{ }nm

m{m^3}