Home
Class 12
CHEMISTRY
Let the solubilities of AgCI in H(2)O, a...

Let the solubilities of `AgCI` in `H_(2)O`, and in `0.01M CaCI_(2), 0.01M NaCI`, and `0.05M AgNO_(3)` be `S_(1),S_(2),S_(3),S_(4)`, respectively. What is the correct relationship between these quantites.

A

`S_(1)gtS_(2)gtS_(3)S_(4)`

B

`S_(1)gtS_(2)=S_(3)gtS_(4)`

C

`S_(1)gtS_(3)gtS_(2)S_(4)`

D

`S_(4)gtS_(2)gtS_(3) gt S_(1)`

Text Solution

Verified by Experts

Promotional Banner

Similar Questions

Explore conceptually related problems

Let the solubilities of AgBr in water and in 0.01M CaBr_(2), 0.01M KBr , and 0.05M AgNO_(3) be S_(1), S_(2),S_(3) and S_(4) , respectively. Give the decreasing order of the solubilities.

The solubility of silver benozate (PhCOOAg) in H_(2)O and in a buffer solution of pH = 4,5 and 6 are S_(1),S_(2),S_(3) , and S_(4) , respectively. Given the decreasing order of their solubilities.

The correct order of increasing solubility of AgCI in (A) water (B) 0.1 M NaCI (C ) 0.1 M BaCI_2 (D) 0.1 M NH_3 is

The solubility of silver benzoate (C_(6)H_(5)COOAg) in H_(2)O and in a buffer solution of pH = 2, 3 , and 4 are S_(1), S_(2), S_(3) and S_(4) respectively. The decreasing order of solubility is

1 L of 0.1 M NaOH, 1 L of 0.1 M KOH , and 2 L of 0.05 M Ba (OH)_(2) are mixed togther. What is the final concentration of the solution.

If solubility of AgCl(s) in water is 10^-5 M then the solubility of AgCl in 0.1 M AgNO_3 will be

Let each of the circles, S_(1)=x^(2)+y^(2)+4y-1=0 , S_(2)=x^(2)+y^(2)+6x+y+8=0 , S_(3)=x^(2)+y^(2)-4x-4y-37=0 touches the other two. Let P_(1), P_(2), P_(3) be the points of contact of S_(1) and S_(2), S_(2) and S_(3), S_(3) and S_(1) respectively and C_(1), C_(2), C_(3) be the centres of S_(1), S_(2), S_(3) respectively. Q. The co-ordinates of P_(1) are :

0.1 M H_(2)S has K_(1)=10^(-5) & K_(2)=1.5xx10^(-12) . What will be the concentration of S^(-2) in the solution.

The p^(H) of 0.1 M NaCI solution is

Calculate the solubility of CoS in 0.1M H_(2)S and 0.15M H_(3)O^(oplus) (K_(sp) of CoS = 3 xx 10^(-26)) . (K_(1) xx K_(2) (H_(2)S) = 10^(-21))