Numerical on Ionic Equilibria

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How much (volume) of 0.001 M HCl(aq.) should be added to 10cm³ of 0.001 M NaOH to change its pH by one unit?

#Chemistry #Physical Chemistry

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5 Answers

1

Kumar Saurabh ·2008-10-30 23:44:27

let it be V
Initially pH=11
final pH=10
so [OH^-]=10^-4 M

so 0.001(10-V)/(10+V)=0.0001
i.e. (10+V)/(10-V)=10

11V=90
V=90/11 cm³

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33

Abhishek Priyam ·2008-10-31 00:02:45

I think kumar is right.....

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Nitin Ramnath ·2008-10-31 00:34:02

I didnt understand the step

0.001(10-V)/(10+V)=0.0001

What is 10-V??

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1

Kumar Saurabh ·2008-10-31 00:58:33

millimoles of NaOH=MV=10*0.001
millimoles of HCl added=V*0.001
millimoles of NaOH neutralized by HCl=V*0.001
millimoles of NaOH left=10*0.001-V*0.001=0.001*(10-V)
total volume=10+V
so [NaOH]=[OH^-]=0.001*(10-V)/(10+V)

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Anuj ·2009-04-10 12:36:07

use m1v1-m2v2=mv

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