41.6mL of an HClO2 solution were titrated with 14.85mL of a .146M NaOH solution to reach the equivalence point, what is the molarity of the HClO2 solution?

Question

I've been stuck on this question for a while now and I keep getting the answer of .0521 but my online tutor is saying it's wrong and I don't know what exactly it is that I'm doing wrong. Please help!!

Answer 1

Hi Saphire, I calculated the same answer you have.

(M for HClO₂)x41.6mL = .146Mx14.85mL

(M for HClO₂) = .0521M

Ask your online tutor to show his work to you!

Answer 2

Good evening. The first thing is to look at the hydronium/hydroxide ratio. Since each molecule will yield 1 hydronium or 1 hydroxide ion, you have a 1:1 molar ratio. At this point you should be able to use C1V1=C2V2.

 

While not a dilution problem specifically, the principle is similar; you need to get the same number of molecules of each for neutralization. You need to convert ml to L so units can cancel.

 

C1x.416L=0.146Mx.1485L

 

C1x0.416L=0.021681moles

 

Divide both sides by the 0.416L

 

C1= 0.05212M

 

You do have only 3 sig figs, so you have the correct number of 0.0521 since the fist zeros are not sig figs, but placeholders. A couple things to consider:

 

Technically, you need units, so it should be 0.0521M

Make sure you have your acid formula correct. Make sure it is not H2ClO2 or such.

Solubility should not be an issue as NaOH is a strong base, so should utilize all of the weak acid and drive to completion, but be sure to verify (check your Ka value)

 

Hope this helps

 

Mike