Chemistry question.. helppp?

Question

A solution with a total volume of 500.00 mL is prepared by diluting 45.0 mL of glacial acetic acid with water. Calculate the [H+] and the pH of this solution. Assume that glacial acetic acid is pure liquid acetic acid with a density of 1.05 g/mL.

AHH!

Answer 1

Okay, you have 45 mL of pure acetic acid. If the density is 1.05 g/mL, that corresponds to 45*1.05 = 47.25 grams. The molecular weight of acetic acid is 60.05, so you have 47.25/60.05 = 0.787 moles of acetic acid.

If acetic acid were a strong acid, you could calculate the [H+] concentration/pH by assuming all of the acid dissociates, thus matching the molarity of the solution (0.787/0.5 = 1.574 M). However, in this case, that's not true, acetic acid is a weak acid with a pKa of 4.76 (at 25 deg. C).

So, you calculate the concentration by assuming that all of the hydronium in solution comes from the acetic acid, and that the concentration of acetic acid doesn't change appreciably due to dissociation.

So in terms of equilibrium equation [A-][H+]/[HA] ;

We know [A-] = [H+] (so the hydronium concentration equals that of the conjugate base concentration);

and that [HA] = 1.574M (approximately, it is only very slightly less)

we know from the pKa value that:

-log(10) ([A-][H+]/[HA]) = 4.76;

thus 10^(-4.76) * [HA] = [A-][H+]

thus 2.735*10^-5 = [A-][H+] = [H+]^2

Thus [H+] = 0.00523 ;

pH = -log(10) (0.00523) = 2.28

So, [H+] = 0.00523 and pH = 2.28

Yay wasn't that fun! ;)

Answer 2

all u should do is to calculate the molarity of glacial acetic acid(molarity of [H+])
45ml gaa(glacial acetic acid)*(1.05g/ml gaa)*(1mol gaa/MW of gaa) / (500ml soln)=?
the answer of above is the molarity of gaa (mol/ml) and =[H+]
so pH=-log[H+]
it's all...