logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  JEEMAIN and AIPMT  >>  Chemistry  >>  Equilibrium
0 votes

The value of $K^{sp}$ of $HgCl_2$ at room temperature is $4 \times 10^{-15}$. The concentration of $Cl^-$ ion in its aqueous solution at saturation point is

(a) $1 \times 10^{-5}$

(b) $2 \times 10^{-5}$

(c) $2 \times 10^{-15}$

(d) $8 \times 10^{-15}$

Can you answer this question?
 
 

1 Answer

0 votes
Answer: $1 \times 10^{-5}$ mol $L^{-1}$
 
$HgCl_2 \rightleftharpoons Hg^{2+} + 2Cl^-$
 
If solubility is X $mol L^-$ then $[Cl^-] = 2X$
 
Now, $4X^3 = K_{sp}$ = solubility product
 
$X = [\frac{K_{sp}}{4}]^{\frac{1}{3}} = [\frac{4 \times 10^{-15}}{4}]^{\frac{1}{3}} = 1 \times 10^{-5}$ mol $L^{-1}$
or $2X = 2 \times 10^{-5}$ mol $L^{-1}$

 

answered Nov 29, 2013 by mosymeow_1
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...