molarity and acid base calcs

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molarity and acid base calcs

Postby juliasloan_4g » Fri Dec 07, 2018 10:50 pm

#6B3 in the textbook edition seven asks about calculating the difference between 200 ml of 0.025M HCL solution and accidentally using a 250 ml flask to make this solution. I know that HCl completely dissociates so calculations with ph can be calculated with the 0.025M. However, i dont understand the solution where for 200 ml you simply take -log(0.025) whileas for 250 the calc is -log((200*0.025)/250)). Could someone clarify the second calculation and why you multiply by 200 and divide by 250

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Re: molarity and acid base calcs

Postby Chem_Mod » Sat Dec 08, 2018 9:43 am

The problem asks for the new pH that was diluted. As such, you need to calculate the new concentration to plug into -log. The calculation inside the log is m1v1=m2v2 rearranged.

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