'V' ml of $$0.25$$ M $$AgNO3$$ required for complete precipitation of chloride ions present in $$50$$ ml of $$0.02$$ M solution of [$$Cr(H2O)6$$]$$Cl3$$.'V' $$=$$ $$-----------------------------------$$ ml

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No. of milli moles of complex $$=$$ $$50$$ ×$$0.02$$ $$=$$ $$1$$[$$Cr(H2O)6$$]$$Cl3$$ ⇌[$$Cr(H2O)6$$]$$+3$$ $$+$$ $$3$$ $$Cl--No$$. of milli moles of $$Cl--$$ $$=$$ No . of  milli moles of $$Ag+$$ $$=$$ $$1$$×$$3$$ $$=$$ $$3$$ $$($$ $$1$$ mole of complex gives $$3$$ mole of $$Cl-)V$$ (ml) ×$$0.25$$ $$=$$ $$3V(ml)$$ $$=30.25$$ $$=$$ $$12$$

Bonding between metal and ligand

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'V' ml of 0.25 M AgNO₃ required for complete precipitation of chloride ions present in 50 ml of 0.02 M solution of [Cr(H₂O)₆]Cl₃.'V' = ----------------------------------- ml

No. of milli moles of complex = 50 $\times$ 0.02 = 1
[Cr(H₂O)₆]Cl₃ $⇌$ [Cr(H₂O)₆]⁺³ + 3 Cl^--
No. of milli moles of Cl^-- = No . of milli moles of Ag⁺ = 1 $\times$ 3 = 3 ( 1 mole of complex gives 3 mole of Cl^-)
V (ml) $\times$ 0.25 = 3
V(ml) $= \frac{3}{0.25}$ = 12

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