Chemistry question (1 Viewer)

CM_Tutor

Moderator
Moderator
Joined
Mar 11, 2004
Messages
2,644
Gender
Male
HSC
N/A
Barium nitrate is Ba(NO3)2 so a 0.0500 M solution has [Ba2+] = 0.0500 M and [NO3-] = 0.100 M.

Sodium hydroxide is NaOH so a 0.100 M solution has [Na+] = 0.100 M and [OH-] = 0.100 M.

100 mL of each solution is combined to a 200 mL solution, so using the dilution formula C1V1 = C2V2 to find the diluted concentrations (C2) from the starting concentrations (C1) with V1 = 100 mL and V2 = 200 mL gives:

[Ba2+] = 0.0500 x 100 / 200 = 0.0250 M

and

[OH-] = 0.100 x 100 / 200 = 0.0500 M.

Now, the solubility reaction that we are looking at is

Ba(NO3)2(s) <----> Ba2+(aq) + 2OH-(aq)

and so

Q = [Ba2+][OH-]2 = 0.0250 x 0.05002 = 6.25 x 10-5.

Ksp(Ba(OH)2) = 2.55 x 10-4 > Q

Since Q < Ksp, according to the dissolution equation, more barium hydroxide would need to dissolve to reach saturation, and so no precipitate will form.
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top