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Topic: Dilution of a Base -> changes pH??  (Read 8024 times)

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Beetle

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Dilution of a Base -> changes pH??
« on: May 31, 2005, 05:49:57 AM »
Hey,

I've got a chem problem and it asks me to dilute 25mL of .1M NaOH until it has a pH of 9.01 and to work out the new volume of base. I keep coming up with the an answer of about 244 L does this seem reasonable?

(actual Question is: What volume of H20 must be added to 25mL of .1M NaOH to produce a solution with pH 9.01?)

Any help would be greatly appreciated

Thanx :)
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Offline AWK

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Re:Dilution of a Base -> changes pH??
« Reply #1 on: May 31, 2005, 06:10:32 AM »
pOH=4.99
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Beetle

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Re:Dilution of a Base -> changes pH??
« Reply #2 on: May 31, 2005, 06:30:26 AM »
pOH=-log[OH-]
therefore
10^-4.99=[OH-]=1.02e-5moles/litre

in 25mL there is .1 x .025 moles of NaOH = 2.5e-3
therefore for [OH-] to be 1.02e-5 the total vol must be:

2.5e-3/1.02e-5 = 244.3Litres


Is that close to correct??

Thank you for your help


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Offline AWK

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Re:Dilution of a Base -> changes pH??
« Reply #3 on: May 31, 2005, 08:12:05 AM »
It is better to use a formula c1 x V1 = c2 x V2
where V2 is a final volume. Volume of water = V2 - V1

Your result is just the final volume. Of course, taking into account significant figures, 25 mL does not matter.
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Offline Borek

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Re:Dilution of a Base -> changes pH??
« Reply #4 on: May 31, 2005, 08:14:45 AM »
Quote from: Beetle on May 31, 2005, 06:30:26 AM
2.5e-3/1.02e-5 = 244.3Litres

Is that close to correct??

Seems OK to me.
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