[shakajjh]

Hi, I'm having a difficult time with with problem, I'm wondering if someone could give me some tips?



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A student prepares a solution of carbonate-free NaOH. Upon performing a rough titration, the student finds the concentration to be 0.0638 M instead of 0.1000 M. Assuming that the volume of the solution is 950.0 mL, how many grams of sodium hydroxide should be added to the solution to produce a concentration as close as possible the desired 0.1000 M? (Assume that the sodium hydroxide is carbonate-free.)
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I would really appreciate any help.

Here is what I know.  I'm pretty sure that I'm suppose to use M1V1 = M2V2, but I don't know what I could solve for in the M2V2 because the volume and Molarity of product has already been given?  I know that I'm looking at this problem wrong though.

Thanks a lot

[Kryolith]

Welcome to the forum 

In 1 L there should be 0.1 mol of NaOH. Calculate the moles of NaOH in the actual solution. Now you can tell how many moles you need to add. Convert moles to gram and fill up with water to 1 L....

EDIT
If the volume should not change, you know that in 950 mL there should be 0.095 mol NaOH. (Neglecting the increase of volume by adding NaOH)

[shakajjh]

awesome, thats the approach that I needed.  I don't know why a 1000 level Analytical chemistry class is giving me so many problems.  I've taken Ochem1 and I feel that it is a lot easier....


Thank you sir

[Kryolith]

You're welcome. 

[shakajjh]

ok I got the answer, but I'm not certain about the number of sig figs that this problem should have.  Our prof is very strict on sig figs and the hw submission software that we use counts it so much..