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Calculate the ratio of solubility of AgC...

Calculate the ratio of solubility of `AgCl " at " 25 ^(@) C ` in 3.0 M `NH_3 (K_(sp) ` of AgCl in `NH_3 " is " 3.1 xx 10^(-3) ) and " in " H_2O (K_(sp) ` of AgCl in `H_2O " is" 1.8 xx 10^(-10)) . ` If answer is `1.15 xx 10 ^(x) ` then x=________?

Text Solution

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The correct Answer is:
4
` AgCl + 2NH_3 hArr [Af(NH_3) _2 ]^(+) +Cl^(-) , K_1`
` ((x)/(3-2x) ) ^(2) K_1 rArr (x) /( 3-2x) =sqrt( K_1) ~~(x)/(3) `
` AgCl hArr Ag^(+) +Cl^(-) ,K_2`
` (x)/(y) =sqrt(( K_1)/(K_2)) xx 3 =sqrt((3.1xx 10^(-3))/(1.8 xx 10^(-10))) xx 3 =1.2 xx 10^(4) `
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