Discussion Forum

The volume (in mL) of 0.1 M $AgNO_{3}$  required for complete precipitation of chloride ions present in 30 mL of 0.01 M solution of [$Cr(H_{2}O)_{15}Cl]Cl_{2}$, as silver chloride is close to

Answer:

Explanation:

mmol of complex = 30 x 0.01 = 0.3 Also, 1 mole of complex $[Cr(H_{2}O)_{5}Cl]Cl_{2}$ gives only two moles of chloride ion when dissolved in solution

$[Cr(H_{2}O)_{5}Cl)Cl_{2} \rightarrow   [Cr(H_{2}O)_{5}Cl]^{2+}+2Cl^{-}$

$\Rightarrow$  mmol of Cl^- ion produced from its 0.3 mmol = 0.6

Hence, 0.6 mmol of Ag⁺ would be required for precipitition

$\Rightarrow$  0.60 mmol  of $Ag^{+}$ = 0.1M x V (in mL)

$\Rightarrow$     V=6mL