Question

if ksp of hg2i2 is 3.2 p×10^-23 )then its solubility is​

Answer 1

We know that......

For this salt

Ksp = 16s^4

So solubility = 4√3.2×10^-23÷16

Answer 2

If $K_{sp}$ of $Hg_2I_2$ is $3.2\times 10^{-23}$, the solubility is $8\times 10^{-8}$

Explanation:

The equilibrium constant for a solid material dissolving in an aqueous solution is the solubility product constant, $K_{sp}$. It stands for the degree of solute dissolution in solution. A substance's $K_{sp}$ value increases with how soluble it is.

Given $K_{sp}$ of $Hg_2I_2$ is $3.2\times 10^{-23}$

We know:

$Hg_2I_2\rightarrow Hg_2^{+2}+2I^-\\ s\ \ \ \ \ \ \ \ \ \ \ \ \ \ s\ \ \ \ \ \ \ \ 2s\\K_{sp}=[Hg_2^{+2}][I^-]^2\\K_{sp}=(s)(2s)^2\\\Rightarrow K_{sp}=4s^3$

Now, we know s stands for solubility

$\therefore Solubility=\sqrt[3]{\frac{K_{sp}}{4}}\\\Rightarrow Solubility= \sqrt[3]{\frac{3.2\times 10^{-23}}{4}}\\\therefore Solubility=8\times 10^{-3}$

To know more about the solubility constant, click on the links below:

https://brainly.in/question/21537731

https://brainly.in/question/8045554

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