Jun 26, 2013
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Hi All,

I'll cut to the chase. The question reads:

3 6 . What is the final Cl- concentration after you mix 50.00 mL 0.25 M HCl with 25.00 mL 0.50 M NaOH?
A. 0.33 M
B. 0.25M
C. 0.17M
D. 0.15 M

According to TBR, the answer is 0.17M because you treat the NaOH like a pure water dilution.

When I read the question, I thought it was an Acid-Base chemistry question, and since both reactants have the same number of moles, there is no limiting reagent, and hence I thought you would have all NaCl and no Cl-. 0M is clearly not a choice, but I was just wondering for clarity purposes.

Thanks for the help.

Sorry for the long message. Cheers and the best of luck in your endeavors.
 

BerkReviewTeach

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You are correct that the H+ completely consumes the OH-, but the Na+ and Cl- are spectator ions that remain in solution completely independent of the reaction taking place. For Na+ and Cl-, it is a dilution where 50 mL 0.25 M Cl- is diluted to 75 mL using a solution that has no Cl- ions in it, so M1V1 = M2V2 becomes (0.25)(50) = M2(75), leading to M2 = (0.25) (50/75) = 0.1667 M Cl-.

Hope this helps.

As you will discover once moderator stops by, this type of post belongs in the Q&A section. This thread will get moved there, and in the future you should post questions in that section.
 

gettheleadout

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As you will discover once moderator stops by, this type of post belongs in the Q&A section. This thread will get moved there, and in the future you should post questions in that section.
Indeed! And remember your solubility rules OP! Moving.