The solubility of AgCl in 0.1 M NaCl is (Ksp of AgCl = 1.2*10 to the p

1.	The solubility of AgCl in 0.1 M NaCl is (Ksp of AgCl = 1.2*10 to the power -10)
Answer» $AgCl$ $<klux>1</klux>.2\times 10^{-9}moles/L$ Answer: The solubility of is is Explanation: Solubility product is defined as the EQUILIBRIUM constant in which a SOLID ionic COMPOUND is dissolved to produce its IONS in solution. It is represented as $K_{sp}$ The equation for the ionization of the silver chromate is given as: $AgCl\leftrightharpoons Ag^{+}+Cl^{-}$ $NaCl\rightarrow Na^{+}+Cl^{-}$ We are given: Solubility of $AgCl$ = S mol/L By stoichiometry of the reaction: 1 mole of $AgCl$ gives 1 mole of $Ag^{+}$ and 1 mole of $Cl^-$ . When the solubility of $AgCl$ is S moles/liter, then the solubility of $Ag^{+}$ will be S moles\liter and solubility of $Cl^-$ will be S+0.1 moles/liter. in 0.1 M NaCl. $K_{sp}=[Ag^+][Cl^-]$ $1.2\times 10^{-10}=[S][S+0.1]$ $S=1.2\times 10^{-9}moles/L$ $AgCl$ $1.2\times 10^{-9}moles/L$ Hence, the solubility of is

The solubility of AgCl in 0.1 M NaCl is (Ksp of AgCl = 1.2*10 to the power -10)

Answer»

$AgCl$ $<klux>1</klux>.2\times 10^{-9}moles/L$ Answer: The solubility of is is

Explanation:

Solubility product is defined as the EQUILIBRIUM constant in which a SOLID ionic COMPOUND is dissolved to produce its IONS in solution. It is represented as $K_{sp}$

The equation for the ionization of the silver chromate is given as:

$AgCl\leftrightharpoons Ag^{+}+Cl^{-}$

$NaCl\rightarrow Na^{+}+Cl^{-}$

We are given:

Solubility of $AgCl$ = S mol/L

By stoichiometry of the reaction:

1 mole of $AgCl$ gives 1 mole of $Ag^{+}$ and 1 mole of $Cl^-$ .

When the solubility of $AgCl$ is S moles/liter, then the solubility of $Ag^{+}$ will be S moles\liter and solubility of $Cl^-$ will be S+0.1 moles/liter. in 0.1 M NaCl.

$K_{sp}=[Ag^+][Cl^-]$