Acid Base Chem Question

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Sophie522

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If 50 mL of 2.0 M NaOH is mixed with 50 mL of 4.0 M nitric acid, what is the pH of the final solution?

A. 1.00
B. 7.00
C. 14.00
D. 10.00
E. not enough information to solve.

This has to be a simple problem, but I'm having trouble. Please help. Thanks.

Correct Answer: A

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topscore?

without even doing the math you can instantly see that b,c,d are all incorrect.

you have 2M NaOH(strong base) with a 4M HNO3, a strong acid.

B would be correct if it was 2M of each with same volumes, thus neutralizing the reaction.

C would be possible if it was the same concentration and more base was added. Or a higher concentration of

D would be possible of the concentrations were reversed or you had a strong base with a very acid.

here's the math part.

50mL X 2M NaOH = 100mmols of OH-

50ml X 4M HNO3 = 200mmols of H+

200 - 100 = 100 mmols of H+ in excess

[H+] = 100mmols/100ml(total volume) = 0.1M

pH = -log[H+] => pH = -log[0.1] => pH = 1
 
I'm doing this problem so I can get a refresher on acid/base chemistry

[H+] = 100mmols/100ml(total volume) = 0.1M

pH = -log[H+] => pH = -log[0.1] => pH = 1

I've gotten to this point but I am having the same problem you're having.
Your equation here actually equals 1M, not .1M as we would like it to.

Are we both making errors or am I looking at this wrong.
 
I also got 1 M or pH=0. Maybe they just wanted you to reason it out without the math part and didn't even check the answer with math.

Or maybe they wanted the concentration of h.
 
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oh youguys are right my math was off. yesterday was tiring. stressing out as my test is coming up so missed it.

deriving the answer works out best.

but you can use the same idea if for other strong acids/strong base questions.
 
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