What is the ph at the equivalence point in the titration of 0.1 m ch3cooh?
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at the equivalence point [CH3COO-] = 0.1 /2 = 0.05M
CH3COO- + H2O <---> CH3COOH + OH-
Ka = 10^-4.76 = 1.74 x 10^-5 = x^2 / 0.05-x
x = [OH-]= 0.000932
pOH = 3.0
pH = 11
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at the equivalence point [CH3COO-] = 0.1 /2 = 0.05M
CH3COO- + H2O <---> CH3COOH + OH-
Ka = 10^-4.76 = 1.74 x 10^-5 = x^2 / 0.05-x
x = [OH-]= 0.000932
pOH = 3.0
pH = 11
HOPE THIS ANSWER IS HELPFUL FOR U
Plzzzzz mark me brainlist plzz
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