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Question: Predict if there will be any precipitate by mixing 50 mL of \(0.01\) m NaCl and 50 mL of \(0.01 \ma...

Predict if there will be any precipitate by mixing 50 mL of 0.010.01 m NaCl and 50 mL of 0.01MAgNO30.01 \mathrm { M } \mathrm { AgNO } _ { 3 } solution. The solubility product of AgCl is 1.5×10−101.5 \times 10 ^ { - 10 }

A

Since ionic product is greater than solubility product no precipitate will be formed.

B

Since ionic product is lesser than solubility product, precipitation will occur.

C

Since ionic product is greater than solubility product, precipitation will occur.

D

Since ionic product and solubility product are same, precipitation will no occur.

Answer

Since ionic product is greater than solubility product, precipitation will occur.

Explanation

Solution

: NaCl+AgNO3\mathrm { NaCl } + \mathrm { AgNO } _ { 3 } AgCl+NaNO3\mathrm { AgCl } + \mathrm { NaNO } _ { 3 }

[Ag+]=12×10−2M=0.5×10−2M\left[ \mathrm { Ag } ^ { + } \right] = \frac { 1 } { 2 } \times 10 ^ { - 2 } \mathrm { M } = 0.5 \times 10 ^ { - 2 } \mathrm { M }

[Cl−]=12×10−2M=0.5×10−2M\left[ \mathrm { Cl } ^ { - } \right] = \frac { 1 } { 2 } \times 10 ^ { - 2 } \mathrm { M } = 0.5 \times 10 ^ { - 2 } \mathrm { M }

Kip−[Ag+][Cl−]\mathrm { K } _ { \mathrm { ip } } - \left[ \mathrm { Ag } ^ { + } \right] \left[ \mathrm { Cl } ^ { - } \right]

=(0.5×10−2)×(0.5×10−2)= \left( 0.5 \times 10 ^ { - 2 } \right) \times \left( 0.5 \times 10 ^ { - 2 } \right)

=2.5×10−5= 2.5 \times 10 ^ { - 5 }

Kip>Ksp\mathrm { K } _ { \mathrm { ip } } > \mathrm { K } _ { \mathrm { sp } }

When it results in precipitation.