Nominal Concentration

Moderators: Chem_Mod, Chem_Admin

504608290
Posts: 18
Joined: Fri Sep 25, 2015 3:00 am

Nominal Concentration

Postby 504608290 » Thu Dec 03, 2015 12:39 am

For number 12.25, it asks you to calculate the initial molarity of and the molarities of , , and in an aqueous solution that contains .43 g of in .1L of solution. How do you do this problem? In the solutions manual it mentions that = nominal concentration of . What does this mean? I've searched everywhere and the term nominal concentration doesn't seem to exist.

Annie Qing 2F
Posts: 28
Joined: Fri Sep 25, 2015 3:00 am

Re: Nominal Concentration

Postby Annie Qing 2F » Thu Dec 03, 2015 2:17 am

So the dissociation reaction is

First, calculate the initial molarity of :

where 171.342 g/mol is the molar mass of .

Then, since we know completely dissociates, by the dissociation reaction above:
= 0.025 M
= 2(0.025 M) = 0.050 M

Finally, since we know = 0.050 M from above,




I think the solutions manual was simply trying to clarify that putting brackets around a chemical formula denotes the concentration of that molecule or compound.


Hopefully this helps!


Return to “Calculating pH or pOH for Strong & Weak Acids & Bases”

Who is online

Users browsing this forum: No registered users and 1 guest