How do you find the pH and % dissociation for the following question?
Question asked by: lenamore
Asked on: 01 Dec 2008
Comment or provide your answer to this question
PH = 1/2pKa-1/2logc; (pKa = -logKa & c =conc. of soln. in moles/L.)
Therefore, pH = -1/2log(10^-5)-1/2log(0.1)
= 2.5+0.5
= 3.0
Percentage of dissociation = √(Ka/c)
= √{(10^-5)/0.1}
= √(10^-4)
= 10^-2 i.e, 1%
By: psahu63
Date of comment: Mon, Sep 6th 2010
Comments and other answers:
PH = 1/2pKa-1/2logc; (pKa = -logKa & c =conc. of soln. in moles/L.)
Therefore, pH = -1/2log(10^-5)-1/2log(0.1)
= 2.5+0.5
= 3.0
Percentage of dissociation = √(Ka/c)
= √{(10^-5)/0.1}
= √(10^-4)
= 10^-2 i.e, 1%
By: psahu63
Date of comment: Mon, Sep 6th 2010
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dissociation following question
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