33I think kumar is right.....
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?
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5 Answers
1let 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
1I didnt understand the step
0.001(10-V)/(10+V)=0.0001
What is 10-V??
1millimoles 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)
