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Question: What will be the solubility of AgCl in \(0.05\)M NaCl aqueous solution if solubility product of AgCl...

What will be the solubility of AgCl in 0.050.05M NaCl aqueous solution if solubility product of AgCl is NH3NH_{3}

A

3×10−9molL−13 \times 10 ^ { - 9 } \mathrm { molL } ^ { - 1 }

B

0.05molL−10.05 \mathrm { molL } ^ { - 1 }

C

H2OH_{2}O

D

3×109molL−13 \times 10 ^ { 9 } \mathrm { molL } ^ { - 1 }

Answer

3×10−9molL−13 \times 10 ^ { - 9 } \mathrm { molL } ^ { - 1 }

Explanation

Solution

: Ksp=[Ag+][Cl−]K_{sp} = \lbrack Ag^{+}\rbrack\lbrack Cl^{-}\rbrack

[Cl−]=NaCl=0.05M\lbrack Cl^{-}\rbrack = NaCl = 0.05M

Ag+=1.5×10−100.05=3×10−9MAg^{+} = \frac{1.5 \times 10^{- 10}}{0.05} = 3 \times 10^{- 9}M

[Ag+]=\lbrack Ag^{+}\rbrack = solubility =3×10−9M= 3 \times 10^{- 9}M