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How to calculate 10% NH4OH equiv solution of NaOH, taking into consideration extra OH |
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| Aug11-12, 08:38 PM | #1 |
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How to calculate 10% NH4OH equiv solution of NaOH, taking into consideration extra OH
Hi, I'm wanting to make a solution of NaOH that has the same neutralizing power per quantity as 10% w/w NH4OH. Given that a 10% solution of both will have an equal amount of dissociated OH, but that NaOH will release additional OH as it is consumed, how should I go about this? Thanks.
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| Aug13-12, 11:11 PM | #2 |
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I'm not quite sure what you mean by "a 10% solution." In order to answer this question, you must first know the molarity of the solutions you want to use.
This is an acid dissociation problem, usually covered in general chemistry 3. If you're not familiar with dissociation constants and equilibrium, it could be a bit confusing for you. |
| Aug14-12, 06:08 PM | #3 |
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10% NH4OH, 90% water, w/w. I'd like to know how to make a solution of NaOH that has comparable neutralizing power, without resulting in PH spikes from additional OH that is released.
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| Aug15-12, 02:14 AM | #4 |
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How to calculate 10% NH4OH equiv solution of NaOH, taking into consideration extra OH
No problem with preparing solution able to neutralize identical amount of acid (simple stoichiometry and concentration conversion problem, similar to the ones explained here: http://www.titrations.info/titration-calculation, you just need to convert molarity to percent concentration), but you can't expect identical pH, it will be much higher in the NaOH solution.
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| Aug15-12, 10:06 PM | #5 |
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