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The molar solubility of AgCl in 1.8 M Ag...

The molar solubility of AgCl in `1.8 M AgNO_(3)` solution is (`K_(sp)` of `AgCl = 1.8 xx 10^(-10)`)

A

`10^(-5)`

B

`10^(-10)`

C

`1.8x10^(-5)`

D

`1.8x10^(-10)`

Text Solution

Verified by Experts

The correct Answer is:
B
`1.8 M AgNO_(3)....... 1.8x10^(-10)`
`1M AgNO_(3)........? `
Molar solublity `=(1.8xx10^(-10))/(1.8)=10^(-10)`
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NARAYNA-ACIDS & BASES-EXERCISE-II (H.W.) (SOLUBILITY PRODUCT & COMMON ION EFFECT)
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