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Calculate the dissociation constant at the same temperaturen given that the PH of a 0.001 monobasic acid is 4.22 at 25 degrees Celsius?
Wiki User
∙ 13y ago
Ka= (x)^2/(.001)
X=[H3O+]=10^-PH=10^(-4.22)=6.02^-5
ka=(6.02^-5)^2/(.001)=3.63^-6
Wiki User
∙ 13y ago
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