Question

A complex of cobalt with ammonia is analyzed for determining its formula, by titrating it against a strandarized acid as follows :

$\mathrm{Co}\left(\mathrm{NH}_3\right)_{\mathrm{x}} \mathrm{Cl}_{\mathrm{y}}$ (aq) $+\mathrm{HCl} \rightarrow \mathrm{NH}_4^{+}$(aq) $+\mathrm{Co}^{\mathrm{y}+}+(\mathrm{aq})+\mathrm{Cl}^{-}$(aq) $\mathrm{A} 1.8 \mathrm{~g}$ A 1.8 g complex required 20.00 mL 1.54 M HCl to reach the equivalence point. Also, if the reaction mixture at equivalence point is treated with excess of AgNO₃ solution, 7.735 g of AgCl precipitate was produced. What is the formula of this complex?

[Given: atomic weight of CO = 59 gmol^-1]

• A. $\mathrm{Co}\left(\mathrm{NH}_3\right)_4 \mathrm{Cl}_3$
• B. $\mathrm{Co}\left(\mathrm{NH}_3\right)_4 \mathrm{Cl}_3$
• C. $\mathrm{Co}\left(\mathrm{NH}_4\right)_4 \mathrm{Cl}_3$
• D. $\mathrm{Co}\left(\mathrm{NH}_3\right)_3 \mathrm{Cl}_4$

Answer 1

Answer: (A)

$\frac{1 \text{.} 8 \text{x}}{\text{M}} = 2 0 \times 1 \text{.} 5 4 \times 1 0^{- 3} = 3 0 \text{.} 8 \times 1 0^{- 3}$ ........(I)

$\frac{1 \cdot 8 \left(\text{x} + \text{y}\right)}{\text{M}} = \frac{7 \text{.} 7 3 5}{1 4 3 \text{.} 5} = 5 3 \text{.} 9 \times 1 0^{- 3}$ ........(II)

$\frac{\text{II}}{\text{I}} ⇒ \, \frac{5 3 \text{.} 9 \times 1 0^{- 3}}{3 0 \text{.} 8 \times 1 0^{- 3}} = 1 \text{.} 7 5 = \frac{\text{x} + \text{y}}{\text{x}} = 1 + \frac{\text{y}}{\text{x}}$

$\frac{\text{y}}{\text{x}} = 0 \text{.} 7 5$

Substituting in (I) $\Rightarrow $

$\frac{1 \text{.} 8 \text{x}}{5 9 + 1 7 \text{x} + 3 5 \text{.} 5 \times \text{0} \text{.} 7 5 \text{x}} = 3 0 \text{.} 8 \times 1 0^{- 3}$

$\therefore $ x = 3.98 = 4 $\therefore $ y = 3