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# Unknown Molarity of Liquid

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Joined: Mar 2007
Unknown Molarity of Liquid

Now this sounds easy, but It becomes much harder when you understand the circumstances...

I have four liquids that can be any of the following:

1.00M HCl
.500M HCl
.100M HCl

1. M NaOH
.5 M NaOH
.1 M NaOH

and the acids have phenolphthalein in them....the only problem is that I need to determine the molarity of these four unknowns using just the four unknowns....

This means I cannot use a pH meter, nor a known Acid/Base to determine neutralization...This is confusing...and if anyone can help me in determining the correct procedure into find the right placement, I would be greatly happy.

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Joined: Mar 2005

I'm going to tentatively say that you need to do titration, but let me look up some stuff in my books/labs before I answer this fully.

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Joined: Mar 2007

you do not need to use titration as you are told the actual concentrations. You just need to find out which is which. This is a logic problem rather than a chemical one.

1. If an excess of base is present with phenolphthalein then the solution turns red.

You only have four liquids A,B,C and D

Set up a series of test tubes tray with 1 cm3 of A, B and C liquid
add 12cm3 (an excess) of D to each. If none turn red then D is an acid

Repeat using B,C and D and adding excess A

Carry on until you find which ones are the acids and the bases.

Now you have to find out the molarity of that acids and the bases.

If the same volume is needed for neutralisation of two samples then they have the same molarity.
If twice as much is needed then the ratio is 2:1
If 10x as much is needed then the ratio is 1:0.1

It's all a bit long winded and there is probably an easier way to do it...

Let me ponder a little

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Joined: Sep 2005

I agree with charco. That sounds like the most rasonable way to work through the problem.

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