1. Assume the range of concentrations for the unknown acids is between 0.30 M an

Question

1. Assume the range of concentrations for the unknown acids is between 0.30 M and 0.80 M and that you will be titrating a 5-mL aliquot of an unknown acid with 0.250 M NaOH.

a. What equivalence point volume (in mL) of NaOH would you expect if you were assigned the most dilute unknown acid?

b. What equivalence point volume (in mL) of NaOH would you expect if you were assigned the most concentrated unknown acid?

2. Calculate the pH of a 25.00 mL sample of 0.500 M cinnamic acid solution. The pKa value for cinnamic acid is 4.45

Explanation / Answer

a) The most dilute means concentration of acid = 0.03 M

So according to equation V1 M1 = V2 M2.

V1 = 5 ml M1 = 0.30 M V2 = ? M2 = 0.250M

Volume of NaOH = 5 × 0.30 /0.250 = 6 ml

b)

The most concentrated means concentration of acid = 0.80 M

So according to equation V1 M1 = V2 M2.

V1 = 5 ml M1 = 0.80 M V2 = ? M2 = 0.250M

Volume of NaOH = 5 × 0.80 /0.250 = 16 ml

c)

C6H5CHCHCOOH <==> C6H5CHCHCOO^- + H^+

0.5 -x x x

pKa = 4.45

Ka = 10^(-4.45) =3.16 × 10^-5

Ka =x^2/(0.5-x)

x is tooo snsmall compared to 0.5 so neglect xix in denominator.

3.16 × 10^-5 × 0.5 = x^2

x^2 =0.0000158 = 1.58 × 10^-5

x = 3.98 × 10^-3

[H^+] = x = 3.98 × 10^-3

pH = -log[H^+] = - log 3.98 × 10^-3 = 2.4

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