..............[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. This doesn't look like an 8th grade question to me.
Determine the [OH−] of a solution that is 0.130 M 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−.
8 answers
I am still very confused
how do you plug the E line into the Kb expression
You're making this too complicated. Plug the E line into 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 the E line 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.
..............[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 the E line 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 worked out 1.8 x 10^-4 = (x)(x)/(0.130-x)
then put the answer -log(0.00474819) into a calculator
an got 2.32347191 but it says that it is wrong
then put the answer -log(0.00474819) into a calculator
an got 2.32347191 but it says that it is wrong
the answer actually might be wrong because I used my answer 2.32347191
to find the ph , and it was right I did
ph=14-2.32347191
ph=11.67652809
to find the ph , and it was right I did
ph=14-2.32347191
ph=11.67652809
I don't know where you got that funny -log number. If you solve
1.8E-4 = (x)(x)/(0.130-x).
Using the quadratic formula I obtained 0.00483 M for (OH^-)
Not using it (using the shortcut) I obtained 0.00482 for (OH^-).
Then pOH is 2.32 and pH is 11.7
1.8E-4 = (x)(x)/(0.130-x).
Using the quadratic formula I obtained 0.00483 M for (OH^-)
Not using it (using the shortcut) I obtained 0.00482 for (OH^-).
Then pOH is 2.32 and pH is 11.7
mastering chemistry was wrong because I put 2.32 in and it was wrong, but when I put 11.7 in it was right
1 answer