Approximately how much water should be added to 10.0 mL of 12.0 M HCl so that it has the same pH as .90 M acetic acid(Ka= 1.8 x 10^-5)?

Answer 1

Um , what grade chemistry is this for ?

Answer 2

first find the [ace acid] (which is 0.0040~M)

use Ka constant formula to solve for [H3o+]

then use C1V1 = C2V2
where c1= [ace acid] = 0.004~M
c2= [initial HCl] = 12M
v1= (initial volume of HCl) = 0.01L
and you have to solve for v2

i think this is right lol

and roman, it's a Grade 12 + level question

Answer 3

since everything is a 1:1 ratio, you know that [acetic acid] = [H3O+]

forgot to add that..

Answer 4

wow i'm just forgetting every thing today:

when you find v2, don't forget to minus the initial volume!!! =) (since they are asking how much is added)

i got 29.80L

Answer 5

Calculate the (H^+) for 0.9 M acetic acid.
Then 10.0 mLof 12M x 12MHCl = ?mLH2O x Mfrom acetic acid.
Solve for mL H2O ahd subtract 10 mL from that to find how much to add.

Answer 6

i want to find the density of acetic acid at 30 degree