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.

Expert's answer

CH_{3}COOH + NaOH = CH_{3}COONa + H_{2}O

C_{1}V_{1} = C_{2}V_{2}

20 cm^{3} x V_{1} = 0.1 mol/L x 21.0 cm^{3}

V_{1} = 0.105 mol/L

At half equivalence

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

n(CH_{3}COOH) = 0.02 L x 0.105 mol/L = 0.0021 mol

n(CH_{3}COOH)_{h.e.} = 0.0021 mol - 0.00105 mol = 0.00105 mol

C(CH_{3}COOH)_{h.e.} = 0.00105 mol (0.0105 L + 0.02 L) = 0.0344 mol/L

pH_{weak acid} = 1/2pK_{a} - 1/2logC = 1/2 x 4.76 - 1/2 x log(0.0344) = 5.5

Learn more about our help with Assignments: Inorganic Chemistry

## Comments

## Leave a comment