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Fed [463]
3 years ago
6

Sufficient strong acid is added to a solution containing na2hpo4 to neutralize one-half of it. what will be the ph of this solut

ion?
Chemistry
1 answer:
Ivahew [28]3 years ago
3 0
Assuming that the initial pH is pHi.
The initial concentration of [OH-] is
[OH]- = 10^(14 - pHi)

Since half of the Na2HPO4 is neutralized, the concentration of [OH-] will be reduced to half. The new pH will be
[OH-]/2 = 10^(14 - pHi)/2 = 10^(14 - pH)
14 - pHi - log 2 = 14 - pH
In terms of pHi,
pH = pHi + log 2
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