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7 December, 01:54

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

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  1. 7 December, 02:21
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    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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