Question

The solubility product of agbr is 4.9 * x10^-9 the solubility of agbr will be

Answer 1

Answer:- $7.0*10^-^5M$

Solution:- The ionization equation for AgBr is written as:

$AgBr(s)\rightarrow Ag^+(aq)+Br^-(aq)$

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 $s=7.0*10^-^5M$ .