Home
Class 12
CHEMISTRY
In 1L saturated solution of AgCI [K(sp) ...

In 1L saturated solution of `AgCI [K_(sp) (AgCI) = 1.6 xx 10^(-10)], 0.1 mol` of `CuCI [K_(sp)(CuCI) = 1.0 xx 10^(-6)]` is added. The resultant concentration of `Ag^(+)` in the solution is `1.6 xx 10^(-x)`. The value of "x" is.

Text Solution

Verified by Experts

The correct Answer is:
7

If is a case of simultaneous solubility of salts with a common ion. Here, solubility product of CuCl is much greater than of AgCl, it can be assumed that `Cl^(-)` in solution comes mainly from CuCl.
`[Cl^(-)] = sqrt(K_(sp)(CuCl)) = 10^(-3)M`
Now, for `AgCl, K_(sp) = 1.6 xx 10^(-10) = [Ag^(+)][Cl^(-)] = [Ag^(+)] xx 10^(-3)`
Promotional Banner

Similar Questions

Explore conceptually related problems

In 1L saturated solution of AgCI[K_(SP)(AgCI)= 1.6xx10^(-10)], 0.1 mole of CuCI [K_(SP)(CuCI)=1.0xx10^(-6)] is added. The resulrant concentration of Ag^(+) in the solution is 1.6xx10^(-x) . The value of "x" is:

In 1 L saturated solution of AgCl[K_(sp)(AgCl)=1.6xx10^(-19)], 0.1" mol of "CuCl[K_(sp)(CuCl)=1.0xx10^(-6)] is added. The resultant concentration of Ag^(+) in the solution is 1.6xx10^(-x) . The value of ''x'' is

If the solubility of AgCl in 0.1 M NaCl is ( K_(sp) of AgCl = 1.2 xx 10^(-10) )

In a saturated aqueous solution of AgBr, concentration of Ag^(+) ion is 1xx10^(-6) mol L^(-1) it K_(sp) for AgBr is 4xx10^(-13) , then concentration of Br^(-) in solution is

The solubility of AgCl in 0.1 M NaCl is ( K_(sp) of AgCl = 1.2 xx 10^(-10) )

A solution is saturated with respect to SrF_(2) K_(sp) = 7.9 xx 10^(-10) and SrCO_(3), K_(sp) = 7.0 xx 10^(-10) . If the fluoride ion concentration is found to be 4.0 xx 10^(-2)M . What is the concentration of carbonates ions.

The value of K_(sp) for CaF_(2) is 1.7 xx 10^(-10) . If the concentration of NaF is 0.1 M then new solubility of CaF_(2) is