Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

$RH_2$(ion exchange resin)can replace $Ca^{2+}$ in hand water as,$RH_2+Ca^{2+}\rightarrow RCa+ 2H^+$.1 litre of hard water passing through $RH_2$ has PH is 2.Hence hardness in PPM of $Ca^{2+}$ is:


Can you answer this question?

1 Answer

0 votes
$\therefore Ca^{2+}=\large\frac{10^{-2}}{2}$
$\Rightarrow \large\frac{40\times 10^{-2}}{2}$
$\Rightarrow 0.2g/l$
$\therefore$Hardness=$\large\frac{0.2\times 10^6}{10^3}$
$\Rightarrow$ 200PPM
Hence (a) is the correct answer.
answered Jan 9, 2014 by sreemathi.v

Related questions

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