- #1

cvc121

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## Homework Statement

a) Determine the concentration of NaCH

_{3}COO. The volume of HCl used to reach the equivalence point is 25.05 mL, the concentration of HCl is 0.10 M, and the volume of NaCH

_{3}COO solution used is 10.0 mL.

b) Using the initial concentrations of NaCH

_{3}COO and HCl, and a reliable literature value for the pK

_{b}for NaCH

_{3}COO or PK

_{a}for CH

_{3}COOH, predict the theoretical pH of the equivalence point.

## Homework Equations

## The Attempt at a Solution

Here is my reasoning and calculations:

a)

The equivalence point is the point in the titration when the number of moles of a standard solution (titrant) is equal to the number of moles of a solution of unknown concentration (analyte).

Therefore, at the equivalence point:

n

_{titrant}= n

_{analyte}

n

_{HCl}= n

_{NaCH3COO}

moles of HCl = (concentration)(volume)

moles of HCl = (0.10 M)(0.02505 L)

moles of HCl = 2.5 x 10

^{-3}mol

moles of HCl = moles of NaCH3COO = 2.5 x 10

^{-3}molConcentration of NaCH3COO = moles / total volume

Concentration of NaCH3COO = 2.5 x 10-3 mol / (0.01 L + 0.02505 L)

Concentration of NaCH3COO = 0.07 M

b)

A reliable literature value for the pKa for acetic acid is 4.756.

pKa = -log Ka

Ka = 10-pKa

Ka = 10-4.756

Ka = 1.754 x 10-5

At the equivalence point of the titration of a weak base with a strong acid, the general reaction H3O+

*(aq)*+ B

*(aq) --->*BH

^{+}

*(aq)*+ H

_{2}O

*(l)*has gone approximately to completion. Therefore, approximately all sodium acetate (NaCH3COO) will be converted to its conjugate acid, acetic acid (CH3COOH), in a 1:1 molar ratio.

Ka = [CH3COO-][H3O+] / [CH3COOH]

1.754 x 10-5 = x2 / (0.07 M – x)

Ka is very small. Therefore, the x value in the denominator is negligible and can be omitted.

1.754 x 10

^{-5}= x

^{2}/ 0.07 M

x

^{2}= (1.754 x 10

^{-5})(0.07 M)

x

^{2}= 1.228 x 10

^{-6}M

x = 1.108 x 10

^{-3}M

[H

_{3}O

^{+}] = x = 1.108 x 10

^{-3}M

Validity check:

(1.108 x 10

^{-3}M / 0.07 M )(100%) = 1.58%

Theoretical pH = -log[H

_{3}O

^{+}]

Theoretical pH = -log[1.108 x 10

^{-3}]

Theoretical pH = 2.96

Could anyone verify my reasoning and calculations? Thanks. All help is very much appreciated.