chemistry

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−.

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  1. ..............[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.

  2. I am still very confused

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  3. how do you plug the E line into the Kb expression

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  4. 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.

  5. 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

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  6. 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

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  7. 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

  8. mastering chemistry was wrong because I put 2.32 in and it was wrong, but when I put 11.7 in it was right

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