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−.
..............[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.
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.
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
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
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
mastering chemistry was wrong because I put 2.32 in and it was wrong, but when I put 11.7 in it was right
To determine the [OH-] of a solution that is 0.130 M in CO32- (carbonate ion), we need to use the Kb value for CO32-.
Step 1: Write the balanced chemical equation for the reaction of CO32- with water:
CO32- + H2O ⇌ HCO3- + OH-
Step 2: Set up the Kb expression for CO32-:
Kb = [HCO3-][OH-] / [CO32-]
Step 3: Since the concentration of CO32- is 0.130 M, we can assume that the concentration of HCO3- formed is negligible compared to CO32- (assuming small x). Therefore, we can express the equilibrium concentration of CO32- as (0.130 - x).
Step 4: Substitute the values into the Kb expression:
1.8×10^-4 = x(x) / (0.130 - x)
Solve the equation for x using the quadratic formula:
x = [ -b ± sqrt(b^2 - 4ac) ] / 2a
where a = -1, b = 0.13, and c = -1.8×10^-4.
Step 5: Calculate the concentration of OH- by substituting the value of x into the expression for [OH-]:
[OH-] = x
By following these steps, you should be able to determine the [OH-] of the solution.
As for the pH and pOH of the solution, once you have determined the [OH-] of the solution, you can use the following formulas:
pH = -log[H+]
pOH = -log[OH-]
Since pH + pOH = 14 at 25°C, you can calculate the pH by subtracting the pOH from 14.
Using these equations, you can determine the pH and pOH of the solution.