Determine the [OH−] of a solution that is 0.130 M in CO32− (Kb=1.8×10−4).

Determine the [OH−] of a 0.130 M solution in CO32− (Kb=1.8×10−4). Determine the pH of a solution that is 0.130 M in CO32−

Determine the pOH of a solution that is 0.130 M in CO32−. …………..[CO3]^2- + HOH ==> HCO3^- + OH

I…………..0.130……………………………. 0….. ………0

C…………-x……………………… x. ………….x

E………….0.130-x…………………….x……. …….X

Kb = (HCO3^-)(OH^-)/[CO3]^2-

Plug the E line into the Kb expression and solve for x = (OH^-). Then calculate pOH = -log(OH^-)

Use pH + pOH = 14 to solve for pH.

Post your work if you get stuck. Doesn’t sound like an 8th grade question to me. I’m still very confused, how do you plug the E line into the Kb expression You are making this too complicated. Plug the E line into the Kb expression like this.

…………..[CO3]^2- + HOH ==> HCO3^- + OH

I…………..0.130……………………………. 0….. ………0

C…………-x……………………… x. ………….x

E………….0.130-x…………………….x……. …….X

Kb = (HCO3^-)(OH^-)/[CO3]^2-

Plug the E line into the Kb expression and solve for x = (OH^-).

Looking at line E, you see (CO3^2-) = 0.130-x

(HCO3^-) = x

(OH^-) = x

Kb given in the problem is 1.8 x 10^-4

Now plug these numbers into the Kb expression which is

Kb = (HCO3^-)(OH^-)/[CO3]^2-

1.8 x 10^-4 = (x)(x)/(0.130-x) and solve for x. I calculated 1.8 x 10^-4 = (x)(x)/(0.130-x)

then put the answer -log(0.00474819) in a calculator

one got 2.32347191 but it says it’s wrong the answer might actually be wrong because i used my answer 2.32347191

to find the ph, and it was just that I did

pH=14-2.32347191

ph=11.67652809 I don’t know where this funny log number comes from. If you solve

1.8E-4 = (x)(x)/(0.130-x).

Using the quadratic formula I got 0.00483 M for (OH^-)

Not using it (using the shortcut) I got 0.00482 for (OH^-).

Then the pOH is 2.32 and the pH is 11.7. The chemistry fluency was wrong because I put 2.32 and it was wrong, but when I put 11.7 it was right.