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  • Molar solubility of AgCl at 25°C is . Its solubility in 0.05 M

Molar solubility of AgCl at 25°C is1\times 10^{ -5 } mol/l . Its solubility in 0.05 M NaCl solution is
 

 

Option: 1

10^{ -5 } M


Option: 2

2\times 10^{ -5 } M
 


Option: 3

2\times 10^{ -9 } M

 


Option: 4

2\times 10^{ -3 } M


Answers (1)

best_answer

Answer (3)

K_{sp}=s^{2}=\left ( 10^{-5} \right )^{2}=10^{-10}

Now, new solubility of AgCl in 0.05 M NaCl

S^{1\, =}\frac{K_{sp}}{0.05}=\frac{10^{-10}}{0.05}=2 \times 10^{-9}M

Posted by

SANGALDEEP SINGH

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