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Home  >>  JEEMAIN and AIPMT  >>  Chemistry  >>  Equilibrium
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Let the solubilities of AgCl in $H_2O$, 0.01 M $CaCl_2$, 0.01 M NaCl and 0.05 M $AgNO_3$ be $S_1, S_2, S_3 and S_4$ respectively. What is the correct relationship between these quantities?

(a) $S_1 > S_2 > S_3 > S_4$

(b) $S_1 > S_2 = S_3 > S_4$

(c) $S_1 > S_3 > S_2 > S_4$

(d) $S_4 > S_2 > S_3 > S_1$

Can you answer this question?
 
 

1 Answer

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Toolbox:
  • Larger the concentrations of common ion, lesser will be the solubility.
 
Answer: $S_1 > S_3 > S_2 > S_4$
 
The equation representing ionisation of AgCl is
$AgCl \rightleftharpoons Ag^+ + Cl^-$
 
Any of the common ion $Ag^+$ or $Cl^-$ will tend to push the equilibrium in backward direction and thus, decrease the solubility of AgCl. Larger the concentrations of common ion, lesser will be the solubility.
 
Now,
in $H_2O$ $\dots \dots \dots \dots $ no common ion.
in $CaCl_2 (0.01 M) \dots \dots \dots [Cl^-]$ = 0.02 M
in $NaCl (0.01 M) \dots \dots \dots [Cl^-]$ = 0.1 M
in $AgNO_3 (0.05 M) \dots \dots \dots [Ag^+]$ = 0.05 M
 
Thus, Solubility of AgCl should be in water ($S_1$) and it should decrease in NaCl ($S_3$) to $CaCl_2(S_2)$ to $AgNO_3(S_4)$

 

answered Nov 23, 2013 by mosymeow_1
 

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