Question

the solubility of the product of AgBr at a certain temp is 2.5 ×10^-13 . find out solubility of AgBr in gram /L at this temperature​

Answer 1

Answer:

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Answer 2

Given: The solubility of the product of AgBr at a certain temp is $2.5×10^{-13}$.

To find: We have to find the solubility.

Solution:

AgBr dissociate as-

$AgBr\rightarrow Ag^++Br^-$

The solubility product of silver bromide is a multiple of solubility of silver cation and bromine anion.

Let the solubility of silver and bromine ion be x.

So, we can write-

$K_{sp}=x^2$

Where Ksp is the solubility product of AgBr.

Putting the value of solubility product in the above formula we get-

${x}^{2} = 2.5 \times {10}^{ - 13} \\ x = \sqrt{2.5 \times {10}^{ - 13}} \\ x = 5 \times {10}^{ - 7}$

Thus, the solubility of AgBr is $5×10^{-7}$ gram/l.