Question

The solubility product of agbr is 4.9×10−94.9×10−9. The solubility of agbr will be

Answer 1

Solubility Product of AgBr =4 x 10^-9 mol/L So solubility of AgBr= (4 x 10^-9)^0.5 mol/L

( Since it ionises as AgBr--------->Ag+ +Br-)

So solubility of AgBr

= (40 x 10^-10)^0.5

= 6.325 x 10^-5 mol/L

Molecular Weight of AgBr

=(108+79.9)=187.9

So solubility of AgBr

= 6.325 x10 ^-5x 187.9g/L

= 1.188 x 10^-2 g/L.

Answer 2

Answer:

Explanation:

The ionization equation for AgBr is written as:

AgBr(s)---Ag+(aq)+Br-(aq)

If the solubility of AgBr is s then the solubility for each product ion would also be same as there is 1:1 mol ratio. The solubility product, Ksp expression is written as:

Ksp=[Ag+][Br-]

Let's plug in the values in it:

4.9*10^-9=(s)(s)

4.9*10^-9=(s)^2

Taking square root to both sides:

s=7.0*10^-5

So, the solubility of AgBr will be s=7.0*10^-5M .

If you wants it in terms of gram/Liter then we could multiply this by the molar mass of AgBr. Molar mass of AgBr is 187.77 g/mol.

So,

7.0*10^-5 mol/L(187.77g/mol) =1.316*10^-2 g/L