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Radda [10]
3 years ago
6

What is the [OH-] in a solution if the [H*] = 1.2 x 10-3 M?

Chemistry
1 answer:
spayn [35]3 years ago
7 0

We know that [OH⁻] * [H⁺] = 10⁻¹⁴

plugging the value of [H⁺]

[OH⁻] * 1.2 * 10⁻³ = 10⁻¹⁴

[OH⁻] = 10⁻¹⁴ * (10³/1.2)

[OH⁻] = 833.3 * 10⁻¹⁴

[OH⁻] = 8.33 * 10⁻¹²

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In the citric acid cycle, ATP molecules are produced by _____. In the citric acid cycle, ATP molecules are produced by _____. su
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substrate-level phosphorylation

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3 years ago
Is this chemical equation balanced?
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5 0
3 years ago
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What can liquids and gases do that solids cannot?
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I hope this helps

5 0
2 years ago
Calculate the percent ionization of nitrous acid in a solution that is 0.139 M in nitrous acid. The acid dissociation constant o
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Ok first, we have to create a balanced equation for the dissolution of nitrous acid.

HNO2 <-> H(+) + NO2(-)

Next, create an ICE table

           HNO2   <-->  H+        NO2-
[]i        0.139M          0M       0M
Δ[]      -x                   +x         +x
[]f        0.139-x          x           x

Then, using the concentration equation, you get

4.5x10^-4 = [H+][NO2-]/[HNO2]

4.5x10^-4 = x*x / .139 - x

However, because the Ka value for nitrous acid is lower than 10^-3, we can assume the amount it dissociates is negligable, 

assume 0.139-x ≈ 0.139

4.5x10^-4 = x^2/0.139

Then, we solve for x by first multiplying both sides by 0.139 and then taking the square root of both sides.

We get the final concentrations of [H+] and [NO2-] to be x, which equals 0.007M.

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≈5.03% dissociation.
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