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Home Q & A Board Science-Related Homework Help Chemistry (Moderator: Laser_3)
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polishprodigy93 polishprodigy93
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4 years ago
The [OH-] of a solution of 3.41 mol/L sodium carbonate is ? in mol/L?
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#1
4 years ago
Sodium carbonate is a weak acid, so it incompletely disassociates:

Na2CO3 + 2 H2O > H2CO3 + 2 NaOH

If [Na2CO3] = 3.41 mol/L, multiply it by 1 L to get the number of moles = 3.41. It's a 1 to 2 ratio between Na2CO3 and NaOH, so multiply that by 2 = 6.82 moles of NaOH. Divide by 1 L and you have the concentration of OH.
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bio_manbio_man
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4 years ago
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#3
4 years ago
Where did you get 2.1 x 10^-4 as the Kb value for CO32-? I look at the chart in my book but I don't see that value anywhere or am I missing something?
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#4
Educator
4 years ago
I found it online, what does your textbook give as the value? Take a picture if possible and upload it
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#5
4 years ago
Pages 8 and 9 are the charts we refer too
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#6
Educator
4 years ago
According to the chart provided, it has a ka of 4.7 × 10–11. Since this is a base, we need to change this to kb using the formula:

kw = ka × kb

Where kw = 1.0 × 10-14

So:

1.0 × 10-14 / 4.7 × 10–11 = 2.1 × 10–4

This is why it was used in the calculation... Face Screaming in Fear Please mark as solved if content with the answer.
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