The solubility product of silver chloride is `1.8 xx 10^(-10)` at 298 K. The solubility of AgCl in 0.01 M HCl solution in mol `// dm^(3)` is
The solubility product of silver chloride is `1.8 xx 10^(-10)` at 298 K. The solubility of AgCl in 0.01 M HCl solution in mol `// dm^(3)` is
A. `2.4 xx 10^(-9)`
B. `3.6 xx 10^(-8)`
C. `0.9 xx 10^(-10)`
D. `1.8 xx 10^(-8)`
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Correct Answer - A
Let solubility of AgCl in 0.01 M HCl `= x mol L^(-1)`
`:. [Ag^(+)]=x mol L^(-1)` ltbr `[Cl^(-)]=[HCl]=0.01M=10^(-2)M`
`[Cl^(-)]` from AgCl can be neglected ]
`K_(sp)==[Ag^(+)][Cl^(-)]`
`1.8 xx 10^(-10)=x xx 10^(-2) implies x =1.8 xx 10^(-8)`
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