Home
Class 11
CHEMISTRY
The solubility product of AgCl is 1.8xx1...

The solubility product of AgCl is `1.8xx10^(-10)`. Precipitation of AgCl will occur only when equal volumes of solutions of :

A

`10^(-8)"M "Ag^(+)and 10^(-8)"M "Cl^(-)` ions

B

`10^(-3)"M "Ag^(+)and 10^(-3)"M "Cl^(-)` ions

C

`10^(-6)"M "Ag^(+)and 10^(-6)"M "Cl^(-)` ions

D

`10^(-10)"M "Ag^(+)and 10^(-10)"M "Cl^(-)` ions

Text Solution

Verified by Experts

The correct Answer is:
B
For precipitation, ionic product `gt` solubility product. In all other options, ionic product is less than solubility product.
Promotional Banner

Topper's Solved these Questions

  • EQUILIBRIUM

    NCERT FINGERTIPS|Exercise Higher Order Thinking Skills|8 Videos

  • EQUILIBRIUM

    NCERT FINGERTIPS|Exercise NCERT Exemplar|19 Videos

  • EQUILIBRIUM

    NCERT FINGERTIPS|Exercise Buffer Solutions|2 Videos

  • ENVIRONMENTAL CHEMISTRY

    NCERT FINGERTIPS|Exercise Assertion And Reason|15 Videos

  • HYDROCARBONS

    NCERT FINGERTIPS|Exercise Assertion And Reason|15 Videos

Similar Questions

Explore conceptually related problems

The solubility of product (K_(sp)) of AgCl is 1.8 xx 10^(-10) . Precipitation of AgCl will occur only when equal volumes of solutions of

The solubility product of AgCl is 1.56xx10^(-10) find solubility in g/ltr

The solubility product of AgCl is 1.8xx10^(-10) at 18^(@)C . The solubility of AgCl in 0.1 M solution of sodium chloride would be

The solubility product of AgCl is 4.0 xx 10^(-10) at 298 K . The solubility of AgCl in 0.04 M Ca Cl_(2) will be