Question

The solubility product of AgCl in water is
$1.5 \times {10}^{ - 10}$
Then calculate it's solubility in 0.01 M NaCl aqueous solution.

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Answer 1

Answer:

Explanation:

AgCl in water =15*10^-10

And

Nacl in aq solution =0.01M

So

NaCl= Na+ Cl-

AgCl =Ag+ Cl -

Let the solubility of Ag+ ion be x mol/ L.

And solubility of Cl- ions will be due to dissociation of AgCl and NaCl both.

Solubility of Cl- ions due to AgCl will be = x mol/ L

Solubility of Cl- ions due to NaCl will be = 0.01 mol/ L

Total solubility of Cl- ions will be = ( x+0.01) mol/ L

Solubility product of AgCl = 1.5 × 10 -10

x ( x+0.01) = 1.5 × 10 -10 ............... (i)

Here x is near about 1.5 × 10 -10.

Now 0.01 is much much greater than 1.5 × 10 -10. So we can neglect x in comparison to 0.01 in ( x + 0.01) .

( x + 0.01) = 0.01

Putting in (i)

x (0.01) = 1.5 × 10 -10

x = 1.5 × 10 -8 mol/L

Ans is 1.5*10^-8

Hope it helps u

Answer 2

$\textsf{\Large {\underline {Equilibrium}}}$ :

$\textsf {\underline {Solution}}$ :

$\boxed{\mathsf{NaCl\:{\rightarrow{{Na} ^{+} \:+\:{Cl} ^{-}}}}}$

Concentration of $\mathsf{{Cl} ^{-}}$ ion in 0.01 M NaCl solution $\mathsf{{[Cl]} ^{-}}$ = 0.01 M

Let solubility of AgCl in 0.01 M NaCl be $\mathsf{x\:mol\:{L} ^{-1}}$

Now, $\boxed{\mathsf{AgCl(s) \:{\rightleftharpoons{{Ag} ^{+} \:+\:{Cl} ^{-}}} }}$

$\mathsf{[Ag} ^{+}]$ = $\mathsf{x\:mol\:{L} ^{-1}}$

$\mathsf{[Cl}^{-}]$ = $\mathsf{x\:mol\:{L}^{-1}}$

Total $\mathsf{[Cl} ^{-}]$ = 0.01 + x = 0.01 [ x is very small ]

Solubility Product, $\mathsf{K_{sp} \:=\:{[Ag} ^{+}]{[Cl}^{-}]}$ = x × 0.01 = 0.01 x

$\mathsf{0.01\:x\:=\:1.5\:*\:{10}^{-10}}$

$\mathsf{x\:=\:{\dfrac{1.5\:*\:{10}^{-10}}{0.01}}}$

$\boxed{\mathsf{x\:=\:1.5*{10}^{-8}mol\:{L} ^{-1}}}$