Question

the solubility of Hg2I2 in water is 10^-5 M calculate its solubility product?​

Answer 1

Answer:

x^x.y^ys^x+y is a right ans

Answer 2

Solubility product of $Hg2I2$ is given by   $K sp = [Hg^{2+} ][I^{-} ]$ which is $4$×$10^{-5}$

Explanation:

• The solubility product is the solid's solubility which is expressed in units of moles per liter (mol/L),and it  is  known as its molar solubility.
• The solubility constant (Ksp) is also known as the solubility product.
• When a solid substance is dissolved in a liquid or an aqueous solution, the equilibrium constant is used in equations.
• $Hg 2 I 2 ---Hg2^{2+} +2I^{-}$
• Here$Hg = s$ and $2I- = 2s$
• For finding the solubility product $K sp = [Hg^{2+} ][I^{-} ]2$

the solubility of $Hg2I2$ in water is

$Ksp=[s][2s]^{2}$

• $Ksp=4s^{3} =4[10^{-5}$]=$4$×$10^{-5}$