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Chemistry Forums for Students => High School Chemistry Forum => Topic started by: Terry.K on July 15, 2005, 01:48:51 AM

Title: Resulting Molarity
Post by: Terry.K on July 15, 2005, 01:48:51 AM

I don't know the formula of this question...

What is the resulting molarity when 150. mL of 0.350 M NaOH solution is combined with 100. mL of 0.200 M NaOH solution?

In my class friends told me the answer is 0.290 M..

but I don't know how to get it.

Title: Re:Resulting Molarity
Post by: Borek on July 15, 2005, 03:28:46 AM

http://www.chembuddy.com/?left=concentration&right=dilution-mixing (http://www.chembuddy.com/?left=concentration&right=dilution-mixing)

Title: Re:Resulting Molarity
Post by: Qazzian on July 17, 2005, 01:42:52 AM

One way (I'm sure there's more) is to find the number of moles in each sample.

So you take the molarity, (mol/L) and multiply by the volume in litres. Do that for each sample, and add the total number of moles.

Now, take the number of total moles, and divide by the TOTAL volume in litres, and you get the concentration. And yes, it is 0.290 M.

Title: Re:Resulting Molarity
Post by: lemonoman on July 17, 2005, 01:51:45 AM

Now is as good a time as any to complement Borek on his ChemBuddy website, and it's comprehensiveness.

It is also as good a time as any to complement Qazzian, on paraphrasing the ChemBuddy site so that Terry doesn't have to click the link.  Power to the lazy! lol...

It is also important to note that understanding how this works is KEY to doing well on your chemistry tests at school.  It is worth the time to learn it.  Good luck  :)

Title: Re:Resulting Molarity
Post by: jdurg on July 18, 2005, 02:49:51 PM

Yeah, fancy calculators are a god send and a time saver to those of us who have to make these calculations on a daily basis.  However, they should not be used as a substitute to actually knowing how to solve the problem.  That's simply because what happens if you need to figure out a question but don't have access to the calculator?   ;)

Title: Re:Resulting Molarity
Post by: Borek on July 18, 2005, 03:35:03 PM

Quote from: jdurg on July 18, 2005, 02:49:51 PM

(...) fancy calculators (...) should not be used as a substitute to actually knowing how to solve the problem.

That's why I have posted a link not to calculator, but to the lecture on the solution mixing calculations. You know, fishing pole, not a fish ;)

Title: Re:Resulting Molarity
Post by: arnyk on July 18, 2005, 07:40:20 PM

Ah the old saying, "Teach a man to fish and he will give you some fish if he catches some but it really depends on how well you taught him."...or something like that.  ::)

Title: Re:Resulting Molarity
Post by: xiankai on July 18, 2005, 08:02:01 PM

that sounds like the chinese saying, "give a man a fish and he'll live for a day. teach a man how to fish and he'll live forever."