Calculate the molarity of an acetic acid solution HC2H3O2...

Question

if a 50.0 mL sample requires 35.8 of 0.150 M NaOH in a titration?

Answer 1

Find the reaction formula aA + bB = cC + dD

to find the MOLE ratio... i.e. how many HC2H3O2 reacts with NaOH

Then... find the number of moles of NaOH in 35.8ml(cm^3)

which:

concentration (molarity) = moles / volume (dm^3)

conc. = 0.150

volume = 35.8 ml = 0.0358dm^3

moles = 0.150 x 0.0358

Then times this number by whatever the moles ratio is to find the number of moles of HC2H3O2.

Then use...

moles of HC2H3O2 / volume = concentration

Remembering to divide the volume by 1000 as you have it in ml instead of dm^3

Answer 2

First, you need to find out how many moles of titrants are needed.

So you take

35.8mL x (1L/1000mL) = A (this is the L of NaOH, you need it to convert the M to moles)

I dont have a calculator, but you take the number you get from the above

which I will denote as A and...

A x ( 0.150mol NaOH/1L) = B     (remember M is mol/L)  B is the mol of NaOH.

Now, multiply this number by the mole ratio ( from the balanced equation)

B x (1mol CH3COOH/1mol NaOH) = c

The mol NaOH wil cancel out.  This will leave you with the moles of CH3COOH which is denoted by C

Now, all you have to do is take the 50.0mL and convert it to L by

50.0mL x (1L/1000mL)= D

Now you just divide C/D (which is the Moles of Acetic Acid over the L of Acetic Acid) and get the Molarity  of Acetic Acid.  Hope this helps.