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For the measurement of the solubility pr...

For the measurement of the solubility product of AgCl the following cell is constructed :
`Ag|AgCl||KCl (0.1 M)||AgNO_(3) (0.1 M)|Ag`
The emf of the cell is 0.45 volt. In the cell, KCl is dissociated to the extent of 83% and `AgNO_(3)` is dissociated to the extent of 86%. Calculate the solubility product of AgCl at 298 K.

Text Solution

Verified by Experts

The correct Answer is:
`1.735xx10^(-10)`

`E_(cell)=E_(cell)^(@)-0.0591 "log "([Ag^(+)]_("Anode"))/([Ag^(+)]_("Cathode"))`
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The EMF of the cell : Ag|AgCl,0.1 MKCl||0.1 M AgNO_(3)|Ag is 0.45V. 0.1 M KCl is 85% dissociated and 0.1 M AgNO_(3) is 82% dissociated. Calculate the solubility product of AgCl at 25^(@)C .

The emf of the cell, Ag|AgI (0.05) MKI|| (0.05) M AgNO_(3)|Ag is 0.788 volt. Calculate the solubility product of AgI.

Knowledge Check

  • The EMF of the following cell is 0.86 volts Ag|AgNO_(3)(0.0093M)||AgNO_(3)(xM)|Ag . The value of x will be

    A
    82.8M
    B
    2.28M
    C
    0.228M
    D
    1.14M
  • Calculate the emf of the following concentration cell at 25^(@)C : Ag(s)|AgNO_(3) (0.01 M)||AgNO_(3) (0.05 M)|Ag (s)

    A
    `-0.414 V`
    B
    `0.828 V`
    C
    `0.414 V`
    D
    `0.0414 V`
  • for the electrochemical cell: Ag|AgCl(s)|KCl(aq)||AgNO_(3)(aq)|Ag . The overall cell reaction is

    A
    `Ag^(+)+KClrarrAgCl+K^(+)`
    B
    `Ag^(+)+Cl^(-)rarr+(1)/(2)Cl_(2)`
    C
    `AgClrarrAg^(+)+Cl^(-)`
    D
    `Ag^(+)+Cl^(-)rarrAgCl`.
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    The emf of the cell Ag|KI(0.05M)||AgNO_(3)(0.05M)|Ag is 0.788V . Calculate the solubility product of AgI .

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