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Topic: titration  (Read 2944 times)

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Offline dess

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titration
« on: August 06, 2011, 12:32:53 PM »
i'd like to ask if how can you solve a titration problem if there is no given volume?

like for instance, Titration of 0.824 g of KHP required 38.314 g NaOH solution to reach the end point  detected by the phenophthalein indicator. Find the molarity of NaOH.

i have something in my mind but i'm not sure. :)

thank you.

Offline mallesh.jubburu

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Re: titration
« Reply #1 on: August 19, 2011, 03:35:11 AM »
HI,
YOU'R SAYING THAT 0.824 GR OF KHP REQUIRES 38.314 GR OF NAOH SOLUTION...OK

LET ME KNOW..
1-NAOH SOLUTION MEANS WHETER THE SOLVENT IS WATER OR METHANOL.....OR ANY OTHER.
2-IF THE SOLVENT IS WATER MEANS WE SHOULD KNOW THE DENSITY OF THE NAOH SOLUTION WHAT WE ARE USING.

IF WE ASSUME DENSITY OF PURE WATER IS 1.0 GM/ML....
NAOH SOLUTION CONTAINS MORE THAN 1.0 .SO IF WE CALICULATE ON THAT BASIS 0.824 GR OF KHP REQUIRES 38.314 GR OF NAOH SOLUTION S0 EACH ML OF NAOH SOLUTION CONTAINS 0.004 GRAMS PER ML THAT MEANS MOLARITY OF NAOH SOLUTION IS AROUN 0.1M



THANKS.
MALLESH.J

Offline Borek

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Re: titration
« Reply #2 on: August 19, 2011, 04:35:58 AM »
YOU'R SAYING THAT 0.824 GR OF KHP REQUIRES 38.314 GR OF NAOH SOLUTION...OK

Please read forum rules. Don't abuse capital letters.

It is easy to calculate number of moles of NaOH and its mass, that gives percent concentration. Use density tables to find solution density, convert to molarity. Done.
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