Home
Class 12
CHEMISTRY
Knowing that K(sp) for AgCl is 1.0 xx 10...

Knowing that `K_(sp)` for `AgCl` is `1.0 xx 10^(-10)`, calculate `E` for a silver `//` silver chloride electrode immersed in `1.00 M KCl ` at `25^(@)C.E^(c-)._(Ag^(o+)|Ag)=0.799V`.

Text Solution

Verified by Experts

The correct Answer is:
0.282v
Promotional Banner

Similar Questions

Explore conceptually related problems

K_(sp) of AgCl is 1xx10^(-10) . Its solubility in 0.1 M KNO_(3) will be :

What would be the electrode potential of a silver electrode dipped in a saturated solution of AgCl in contact with 0.1 M KCl solution at 25^(@)C ? E^(c-)._(Ag^(o+)|Ag)=0.799 V K_(sp) of AgCl=1xx10^(-10)

Calculate the potential of silver electrode in a saturated solution of AgBr(K_(sp)=6xx10^(-13)) containing 0.1 M KB r . E^(c-)._(Ag^(o+)|Ag)=0.80V.

K_(sp) of AgCl is 2.8 xx 10^(-10) at 25^(@)C . Calculate solubility of AgCl in. a. Pure water b. 0.1M AgNO_(3) c. 0.1M KCl or 0.1M NaCl

Find the solubility of AgCl in 0.1 M CaCl_(2) . E^(c-)._(Ag^(o+)|Ag)=0.799V and that of AgCl|Ag=0.222V .

The solubility product of silver chloride is 1 xx 10^(-10)" at 25"^@C. Calculate the solubility of silver chloride in 0.1 M sodium chloride.

The K_(sp) of AgCl at 25^(@)C is 2.56 xx 10^(-10) . Then how much volume of H_(2)O is required to dissolve 0.01 mole of salt ?

The solubility product of silver chloride is 1.44 xx 10 ^(-4) at 373 K. The solubility of silver chloride in boiling water will be

K_(sp) of AgCl in water at 25^(@)C is 1.8xx10^(-10) . If 10^(-5) mole of Ag^(+) ions are added to this solution. K_(sp) will be:

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