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Reaction of calgon with hard water containing $\mathrm{Ca}^{2+}$ ions produce
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Verified Answer
The correct answer is:
$\left(\mathrm{Na}_2 \mathrm{CaP}_6 \mathrm{O}_{18}\right]^{2-}$
Calgon is a trade name of a complex salt, sodium hexametaphosphate $\left(\mathrm{NaPO}_3\right)$. Calgon ionises to give a complex anion. The addition of calgon to hard water causes the calcium ions of hard water to displace sodium ions from the anion of calgon and form a complex with calgon.
$$
\begin{aligned}
& \left(\mathrm{NaPO}_3\right. \text { b or } \\
& \mathrm{Na}_2\left(\mathrm{Na}_4 \mathrm{P}_{\mathrm{B}} \mathrm{O}_{1 \mathrm{\theta}}\right) \longrightarrow 2 \mathrm{Na}^{+}+\mathrm{Na}_4 \mathrm{P}_6 \mathrm{O}_{1 \mathrm{~B}}^{2-} \\
& \text { (Complex } \\
& \text { anioi) } \\
& \mathrm{Ca}^{2+}+\mathrm{Na}_4 \mathrm{P}_6 \mathrm{O}_{1 \theta}^{2-} \longrightarrow 2 \mathrm{Na}^{+}+\mathrm{CaNa}_2 \mathrm{P}_6 \mathrm{O}_{1 \mathrm{\theta}}^{2-} \\
& \text { From Anion of calgon } \\
& \text { hard } \\
& \text { Goes into } \\
& \text { solution } \\
&
\end{aligned}
$$
$$
\begin{aligned}
& \left(\mathrm{NaPO}_3\right. \text { b or } \\
& \mathrm{Na}_2\left(\mathrm{Na}_4 \mathrm{P}_{\mathrm{B}} \mathrm{O}_{1 \mathrm{\theta}}\right) \longrightarrow 2 \mathrm{Na}^{+}+\mathrm{Na}_4 \mathrm{P}_6 \mathrm{O}_{1 \mathrm{~B}}^{2-} \\
& \text { (Complex } \\
& \text { anioi) } \\
& \mathrm{Ca}^{2+}+\mathrm{Na}_4 \mathrm{P}_6 \mathrm{O}_{1 \theta}^{2-} \longrightarrow 2 \mathrm{Na}^{+}+\mathrm{CaNa}_2 \mathrm{P}_6 \mathrm{O}_{1 \mathrm{\theta}}^{2-} \\
& \text { From Anion of calgon } \\
& \text { hard } \\
& \text { Goes into } \\
& \text { solution } \\
&
\end{aligned}
$$
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