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) $+\mathrm{HCl}\to {\mathrm{NH}}_4^{+}$(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)

$\frac{1\text{.}8\text{x}}{\text{M}}=20\times 1\text{.}54\times 10^{-3}=30\text{.}8\times 10^{-3}$ ........(I)

$\frac{1\cdot 8\left(\text{x}+\text{y}\right)}{\text{M}}=\frac{7\text{.}735}{143\text{.}5}=53\text{.}9\times 10^{-3}$ ........(II)

$\frac{\text{II}}{\text{I}}\Rightarrow \frac{53\text{.}9\times 10^{-3}}{30\text{.}8\times 10^{-3}}=1\text{.}75=\frac{\text{x}+\text{y}}{\text{x}}=1+\frac{\text{y}}{\text{x}}$

$\frac{\text{y}}{\text{x}}=0\text{.}75$

Substituting in (I) $\Rightarrow$

$\frac{1\text{.}8\text{x}}{59+17\text{x}+35\text{.}5\times \text{0}\text{.}75\text{x}}=30\text{.}8\times 10^{-3}$

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