#1
Hey Pit geniuses,

first test in AP Chem tomorrow. I'm trying to grasp most of the stuff and I'm having a lot of trouble with Acid-Base Titrations:

During a titration the data were collected. A 10.0 mL portion of an unknown monoprotic acid solution was titrated with 1.0 NaOH; 40.0 mL of the base were required to neutralize the sample.

A. What is the molarity of the acid solution?

If you guys get this, it would really help. I know the answer is 4.0 M, but I don't know how to get there.
There was a woman who gave out her pearls.
#2
I got 20% on my chemistry test. Sure you want my help?

IMO, go to Yahoo! Answers. You can ask questions like this and there is a section for chemistry. Pretty good set up. They'd be much better help than anyone here.
#4
Quote by beadhangingOne
m1v1 = m2v2

this (40mL)(1)=10mL(m1) m1=4
Quote by nincompoop
potcorn56, you are a god.
Last edited by potcorn56 at Oct 10, 2008,
#5
For NaOH n=(MV)/1000
n=(40*1)/1000= 0.04
Since it's a monoprotic acid, it means the ratio of acid:base is 1:1, so there is 0.04 moles of acid present.

M=(n*1000)/V=(0.04*1000)/10=4M
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