[John_Kimble]

Some monoprotic acid (HA) dissociates (ionizes) 3.5% if 100 g of the acid is dissolved in 1 dm3 of solution. Calculate its molar mass! K(HA) = 6.7 × 10^-4 mol/dm3.



I don't know. Is that K(HA) a mistake? Is that meant to be concentration or an equilibrium constant? It has a unit, so it should be concentration, but these professors are using units for constants elsewhere as well

So my thinking was... molar mass = mass (100g) / number of moles. I know the percent ionization and the initial concentration of HA (6.7 x 10^-4 mol/dm3 I guess?), so I can calculate the concentration of H+, but what good would that do?

Seems easy enough, but I just don't know   

Thanks to anyone who'll help.

[Borek]

I suppose it is the acid dissociation constant. See if you can use percent dissociation and Ka to calculate molar concentration of the acid.

[John_Kimble]

Ok, I got this for now:


HA    A^-   +   H⁺

Ka = [H+] [A-]  /  [HA]

0,035 = [H+]  /  [HA]initial


Ka is given, and I guess [H+] = [A-] ? But that still leaves me with 3 unknowns.

[Borek]

Think in terms of mass balance. Sum [HA]+[A^-] is a constant.

[John_Kimble]

No, no idea  This one's suppose to be real easy too, since it's the first one.

The correct result is said to be 189.47 g/mol.

[Borek]

This is a simple stoichiometry. You start with C_HA (which you called [HA]initial). Some of the HA dissociates, producing equimolar amount of A^-. That means [HA]+[A^-] doesn't change - HA that dissociates is replaced by A^-. And it should be obvious what [HA]+[A^-] is equal to.

[John_Kimble]

So the K(HA) from the instructions is acid dissociation constant or initial concentration of HA?

[Borek]

IMHO it is just K_a (acid dissociation constant).

[John_Kimble]

Some monoprotic acid (HA) dissociates (ionizes) 3.5% if 100 g of the acid is dissolved in 1 dm3 of solution. Calculate its molar mass! K(HA) = 6.7 × 10^-4 mol/dm3.



So there's an acid, and we throw it into the water, where 3.5% of it will dissociate into A- and H+ ions. The rest (96.5%) will stay as it is. So the reaction for the 3.5% that does dissociate is:

HA   :equil::   A-   +   H+


Now we are given the value of the acid dissociation constant, which is 6.7x10^-4. This value will tell us the ratio between the concentrations of a product of dissociated ions and the concentration of the undissociated acid at equilibrium:

Ka = ( [A-] [H+] ) / [HA]


We can also use the data that 3.5% dissociate, to set up this from the ionization percent equation:

0.035 = [H+] / [HA]initial


The initial concentration of the acid is also the same as the concentration of the undissolved acid at equilibrium and the dissolved one:

[HA]intitial = [HA] + [A-]


Now to get the molar mass of the acid, we need the [HA]initial. The mass is given at 100g.


That's my thinking, but I really don't know how to proceed with this. Please, help me out Borek, so I can have a Christmas dinner, without obsessing over this And merry christmas to you too!

Thanks!

[Borek]

At the moment you have 4 unknowns and three equations - but you have already mentioned fourth equation in your second post.

[John_Kimble]

Oh my god, I got the correct answer Thank you Borek!

However, is this really the easiest procedure?  It's the 1st exercise on the list, and usually those are fairly basic, short and simple.