The solubility product of Hg_2Cl_2 in water a | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The dissociation of $Hg_2Cl_2$ is given by: $Hg_2Cl_2 \rightleftharpoons Hg_2^{2+} + 2Cl^-$Let the solubility of $Hg_2Cl_2$ be $x \, mol/L$.Then,$

Vedclass · Accepted Answer

$2.0 	imes 10^{-6} \, M$

Vedclass · Answer

(C) The dissociation of $Hg_2Cl_2$ is given by: $Hg_2Cl_2 ightleftharpoons Hg_2^{2+} + 2Cl^-$Let the solubility of $Hg_2Cl_2$ be $x \, mol/L$.Then,$[Hg_2^{2+}] = x$ and $[Cl^-] = 2x$.The solubility product expression is: $K_{sp} = [Hg_2^{2+}][Cl^-]^2$.Substituting the values: $K_{sp} = (x)(2x)^2 = 4x^3$.Given $K_{sp} = 3.2 	imes 10^{-17}$,we have:$4x^3 = 3.2 	imes 10^{-17}$$x^3 = 0.8 	imes 10^{-17} = 8 	imes 10^{-18}$$x = \sqrt[3]{8 	imes 10^{-18}} = 2 	imes 10^{-6} \, M$.

The solubility product of $Hg_2Cl_2$ in water at $25 \, ^\circ C$ is $3.2 \times 10^{-17} \, mol^3 \, dm^{-9}$. The solubility of $Hg_2Cl_2$ at $25 \, ^\circ C$ is:

Similar Questions

The solubility of $Ba_3(AsO_4)_2$ (molar mass $= 690 \ g/mol$) is $6.9 \times 10^{-2} \ g/100 \ mL$. What is the value of its $K_{sp}$?

At $25\,^{\circ}C$,the solubility product of $Mg(OH)_2$ is $1.0 \times 10^{-11}$. At which $pH$ will $Mg^{2+}$ ions start precipitating in the form of $Mg(OH)_2$ from a solution of $0.001\, M\, Mg^{2+}$ ions?

Calculate the molar solubility of $Ni(OH)_2$ in $0.10 \ M \ NaOH$. The solubility product of $Ni(OH)_2$ is $2 \times 10^{-15}$.

Which of the following compounds is sparingly soluble in ammonia?

For a sparingly soluble salt $A_pB_q$,the relationship of its solubility product $(L_S)$ with its solubility $(S)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module