Topic: Hydroxide Ion Concentration
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Answers to Common Questions
How to Find Hydroxide Ion Concentration
Distilled water weakly dissociates, forming hydrogen (H+) and hydroxide (OH-) ions (H2O = H+ OH-). At a given temperature, the product of molar concentrations of those ions is always a constant: [H+] x [OH] = constant value. The water ion p... Read More »
Source: http://www.ehow.com/how_5791224_hydroxide-ion-concentration.html
What is the hydroxide ion concentration?
Kw = 10^-13.997 = 1.01 x 10^-14 Kw = [H3O+][OH-] [OH-] = Kw / [H3O+] = 1.01 x 10^-14 / 0.000150 = 6.71 x 10^-11 Or, in your format: [OH-] = 6.71E-11 M Read More »
Source: http://answers.yahoo.com/question/index?qid=20080426154651AAMPRMk
What is a Hydroxide Ion?
A hydroxide ion is one hydrogen atom attached to one oxygen atom. It is an ion because it has a negative charge. Sodium hydroxide is formed when the hydroxide ion combines with the sodium cation (positive charge). They attracted to one anot... Read More »
Source: http://answers.ask.com/Science/Chemistry/what_is_a_hydroxide_ion
More Common Questions
Answers to Other Common Questions
There are two equations to use: pH + pOH = 14.00 [OH-] = 10^-pOH pOH = 14.00-8.06 pOH = 5.94 [OH-] = 10^-5.94 [OH-] = 1.15*10^-6M
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Source: http://answers.yahoo.com/question/index?qid=20101122060815AAoSBoc
You just have to recall that: pH + pOH = 14 [H+][OH-] = 10^-14 pH = -log[H+] pOH = -log[OH-] For example, for the first problem, you're given the pH = 3.00. So the pOH must be 14 - 3.00 = 11.00. And since pOH = -log[OH-], you have 11.00 = -...
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Source: http://answers.yahoo.com/question/index?qid=20081112111141AAir7UD
Moles Ba(OH)2 = 0.200 g / 171.342 g/mol = 0.00117 Concentration Ba(OH)2 = 0.00117 / 0.5 L =0.00234 [OH-] = 2 x 0.00234 = 0.00468 M [H+] = 1.0 x 10^-14 / 0.00468 =2.14 x 10^-12 M
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Source: http://answers.yahoo.com/question/index?qid=20080308053447AAHgGxm
pOH = 6.6 [OH-] = 10^-pOH [OH-] = 10^-6.6 [OH] = 2.5*10^-7 concentration of hydroxide ions = 2.5*10^-7M
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Source: http://answers.yahoo.com/question/index?qid=20090724055254AAzxbX0
Assuming you're working in water at 298K (25 degrees Celsius), pH = - log(10) [H+] [H+] = 10^(-2.3) = 5.01 x 10^-3 mol dm^-3 The ionic dissociation of water (K(w)) is always constant at this temperature, at 1.00 x 10^-14 mol^2 dm^-6. This i...
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Source: http://answers.yahoo.com/question/index?qid=20091203212557AA32Uvq
pOH = 14 - 5.3 = 8.7 [OH-] = 10^- 8.7 =2 x 10^-9 M There is another way : [H+] = 10^-5.3 = 5.0 x 10^-6 M Since [H+][OH-] = 1.0 x 10^-14 [OH-] = 1.0 x 10^-14 / 5.0 x 10^-6 = 2 x 10^-9 in the same way pOH = 14 - 11.9 = 2.1 [OH-] = 10^-2.1 =0....
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Source: http://answers.yahoo.com/question/index?qid=20080803111514AAvUcdX
Since you need the OH- concentration, first find the pOH using the formula pH + pOH = 14.00 7.80 + pOH = 14.00 pOH = 6.20 From pOH you can calculate the [OH-] using the formula 10^-pOH [OH-] = 10^-pOH =10-^-6.20 = 6.3 x 10^-7 M
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Source: http://answers.yahoo.com/question/index?qid=20091128134208AAiN3TE
