Forty milliliters, 40 mL, of 2.0 molar NaOH reacts with 8.0 mL of a monoprotic
acid. What is the molarity of the acid?
a. 0.40 M
b. 2.0 M
c. 10. M
d. 4.0 M

Hello,  
 
This question also follows the same logic as the previous question
that I
answered for you:
http://answers.google.com/answers/main?cmd=threadview&id=227189 
 
Using the equation from that question: 
 
V1 * C1 = V2 * C2 
 
where V1 = volume solution 1    V2 = volume solution 2 
      C1 = concentration solution 1  C2 = concentration solution 2 
 
The acids and bases in this question are both mono-functional (ie they
have one H+ or one OH- group - this is what 'monoprotic' means, it
gives one proton, or one H+) so their acid/base concentration is
given by the concentration of the parent compound.  Thus, plugging in
numbers:
 
V1 = 40 mL  C1 = 2.0M   V2 = 8.0 mL  C2 = x 
 
40 mL * 2.0 M = 8.0 mL * x    and rearranging: 
 
     40 mL * 2.0 M 
x = ----------------   =  10 M monoprotic acid 
        8 mL          
 
 
another description of the titration process, this one with a
flowchart:
http://www.measurementuncertainty.org/mu/guide/index.html?content_frame=/mu/guide/example_a3.html

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titration example