Answer to Question #168709 in Inorganic Chemistry for Arya

Question #168709

20cm^3 of ethanoic acid were titrated with 0.100 moldm^-3 NaOH. At the end point of the reaction 21.0cm^3 of alkali were required.


Calculate the pH at Half equivalence I.e. the pH when 10.5cm^3 of the alkali had been added.


1
Expert's answer
2021-03-08T06:06:14-0500

CH3COOH + NaOH = CH3COONa + H2O

C1V1 = C2V2

20 cm3 x V1 = 0.1 mol/L x 21.0 cm3

V1 = 0.105 mol/L

At half equivalence

n(NaOH) = 0.0105 L x 0.1 mol/L = 0.00105 mol

n(CH3COOH) = 0.02 L x 0.105 mol/L = 0.0021 mol

n(CH3COOH)h.e. = 0.0021 mol - 0.00105 mol = 0.00105 mol

C(CH3COOH)h.e. = 0.00105 mol (0.0105 L + 0.02 L) = 0.0344 mol/L

pHweak acid = 1/2pKa - 1/2logC = 1/2 x 4.76 - 1/2 x log(0.0344) = 5.5


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