Home
Class 11
CHEMISTRY
The solubility of Pb(OH)(2) in water is...

The solubility of `Pb(OH)_(2)` in water is `6.7 xx10^(-6) M` Calculate solubility of `Pb(OH)_(2)` in a buffer of pH=8.

A

`12.03 xx 10 ^(-2)` mol /L

B

`1.203 xx 10 ^(-3)` mol /L

C

`3.102 xx 10 ^(-3)` mol /L

D

`3.102 xx 10 ^(-2) ` mol /L

Text Solution

Verified by Experts

The correct Answer is:
B
`K _(sp ) of P b (OH ) _(2) = 4 S ^(3) = 4 xx ( 6.7 xx 10 ^(-5)) ^(3) = 1. 203 xx 10 ^(-5)` The buffer contains pH = 8
`therefore pOH = 6 or [OH ^(-)] = 10 ^(-5) ,` Now let solubility of `Pb (OH) _(2)` be S mol/L in it Thus
`[Pb ^(2+)] [OH ^(-)] = K _(sp) , [Pb ^(2+)] [10 ^(-6) ] ^(2) = 1. 203 xx 10 ^(-15) [ Pb ^( 2+)] = (1 .203 xx 10 ^(-15))/(10 ^(-12))= 1.203 xx 10 ^(-3)` mol/L
Promotional Banner

Similar Questions

Explore conceptually related problems

The solubility of Mg(OH)_2 at 298K is 1.5xx10^-4 . Calculate the solubility product.

the solubility product of Al(OH)_3 is 1xx10^-36 . Calculate the solubility of Al(OH)_3.

The solubility of BaSO_4 in water at 298K is 1.1 xx10^(-5) mol L^(-1) . Calculate the solubility product of Baso_4 at 298K.

If the solubility of Ag_(2)CrO_(4) in water is C mole/litre, Its solubility product will be

The solubility of H_2S gas in water at STP.195m .Calculate its Henry's law constant at STP

The solubility product of PbCl_2 is 1.68xx10^-6 at 298K. Calculate its solubility at this temperature.

The solubility product of silver chloride is 1.2xx10^(-10) at 298K. Calculate the solubility of AgCI at 298K.