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

$K_{sp}$ of $Mg(OH)_2$ is $1 \times 10^{-12}$, 0.01 M $MG^{2+}$ will precipitate at the limiting $pH$ of

(a) 3

(b) 9

(c) 5

(d) 8

Can you answer this question?
 
 

1 Answer

0 votes
Answer: 9
 
$K_{sp}$ for $Mg(OH)_2 = [Mg^{2+}][OH^-]^2 = 10^{-12}$
$(0.01)(OH^-)^2 = 10^{-12} or [OH^-] = 10^{-5}$
 
$[H^+] = \frac{K_w}{[OH^-]} = \frac{10^{-14}}{10^{-5}} = 10^{-9}$
$pH = - log [H^+] = 9$

 

answered Nov 27, 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
...