With the first pair of solutions, it is not possible to get a neutral solution, since a pH of 7 does not lie anywhere between a pH of 1 and a pH of 4.
It is possible to get a neutral solution with the second pair of solutions, if they are mixed in the appropriate proportions. The simplest case to calculate would be where the first was a very dilute solution of a strong acid and the second a very, very dilute solution of a strong alkali. For instance, one could use 10^(-4)M hydrochloric acid and 10^(-6)M sodium hydroxide. Since the NaOH is 100 times as dilute as the acid, mixing one volume of the acid solution with 100 volumes of the alkaline solution will give a neutral solution.
Of course, if a weak acid and/or a weak alkali are used, or if buffer solutions are used, the calculation becomes much more complicated.
2007-01-12 05:00:15
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answer #1
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answered by deedsallan 3
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It depends of the quantity of each solution. For example, if you put a lot of pH 4 solution into a pH 8 solution you'll get an acid solution, not a neutral. So, you got to put a little pH=4 solution to adjust the pH=8 solution to value 7. You can make it by ear (measuring the pH till you get what U want) or calculate the quantities of each solution.
2007-01-11 09:15:45
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answer #2
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answered by Lu 2
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Since neutral is defined as pH 7, you can not get a pH 7 solution by mixing a pH 1 and pH 4 solution together but you can get a pH 7 solution by mixing a pH 4 and pH 8 solution together.
2007-01-11 07:57:21
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answer #3
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answered by RICHARD M 1
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pH1 & pH4 No!!! (Both too acidic)
pH4 & pH8 Yes!!! (one is acidic and one is alkaline)
Neutrality is pH7
2007-01-14 09:03:30
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answer #4
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answered by lenpol7 7
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No to the first one and yes to the second one. PH7 is neutral
2007-01-11 07:59:46
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answer #5
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answered by Dionysia C 2
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only with the second, but as pH scale is a log scale it is not as easy as a simple average
2007-01-11 08:30:26
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answer #6
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answered by gramps 3
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the first one--no
second one --yes
2007-01-11 07:57:33
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answer #7
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answered by Anonymous
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no
2007-01-11 07:59:31
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answer #8
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answered by Richard Bricker 3
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