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Chemistry Forums for Students => Analytical Chemistry Forum => Topic started by: jkulier on September 25, 2008, 12:45:42 AM

Title: Production of a solution with [Cl-] = 0.02 M
Post by: jkulier on September 25, 2008, 12:45:42 AM

The first part of the question was:

Calculate the mass of the unknown sample needed to produce 40 mL of solution at [Cl−] = 0.02 M.
Assume that the weight percent of NaCl in the unknown is 50%.

For this, I got an answer of 0.09 grams which is correct. However, the next part of the question asks WHY in the previous question, we choose to aim for 0.02 M?

I am drawing up a blank for this, so please point me in the right direction.  ;D

Thank you.

Title: Re: Production of a solution with [Cl-] = 0.02 M
Post by: Borek on September 25, 2008, 02:43:47 AM

Any context to that question?

Title: Re: Production of a solution with [Cl-] = 0.02 M
Post by: jkulier on September 26, 2008, 12:27:56 AM

There actually wasn't any context; this was exactly how it appeared on my homework page. . .

however! In the lab, I will be trying to form the precipitate of AgCl by adding silver nitrate to a solution containing chloride. Ah, also, the silver nitrate concentration will be 0.02M !! Although, I still do not know the answer to the question. . .

thank you again.

Title: Re: Production of a solution with [Cl-] = 0.02 M
Post by: Borek on September 26, 2008, 03:13:46 AM

Quote from: jkulier on September 26, 2008, 12:27:56 AM

however! In the lab, I will be trying to form the precipitate of AgCl by adding silver nitrate to a solution containing chloride. Ah, also, the silver nitrate concentration will be 0.02M !! Although, I still do not know the answer to the question. . .

IMHO that's the context. Could be they want you to be easily able to measure stoichiometric amount of Cl^- without a need of calculation.