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Topic: inorganic  (Read 2967 times)

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Offline snigdha

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inorganic
« on: April 15, 2011, 03:17:13 PM »
What is the difference in pH for 1/3 and 2/3 stages of neutralisation of 0.1 M CH3COOH with 0.1 M NaOH.
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Online Borek

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Re: inorganic
« Reply #1 on: April 15, 2011, 04:17:46 PM »
You have to show your attempts at solving the question to receive help. This is a forum policy.

Please read forum rules.
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Offline snigdha

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Re: inorganic
« Reply #2 on: April 15, 2011, 04:22:48 PM »
im sorry bout that but...
I couldn't figure out the meaning of the question ,i.e the part '1/3 and 2/3 stages of neutralisation'.
so how could i solve it then.. :-\
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Online Borek

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Re: inorganic
« Reply #3 on: April 15, 2011, 06:03:46 PM »
1/3rd (2/3rds) of acetic acid has been neutralized.
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Offline snigdha

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Re: inorganic
« Reply #4 on: April 16, 2011, 03:15:19 AM »
ic..so in this question do we have to use the formula-
pH= -log[H+]?
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Re: inorganic
« Reply #5 on: April 16, 2011, 06:01:10 AM »
No. Hint: you have a solution of a weak acid and its conjugate base.

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Offline snigdha

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Re: inorganic
« Reply #6 on: April 16, 2011, 07:11:56 AM »
oh well then ..
pH=pKa + log[acid]/[conjugate base] ?
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Offline rabolisk

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Re: inorganic
« Reply #7 on: April 16, 2011, 08:08:15 AM »
Check that equation again...
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