olecular wei n 1040615 and Malonic Acid information: the 1.5x10 and Ku 2.0v10\"

Question

olecular wei n 1040615 and Malonic Acid information: the 1.5x10 and Ku 2.0v10" Measure 07 to 0.8 sams of 2. Dissolve your malonic acid in 100 ml of distiled wai alkuot of vour maonic acid solution and put in a 125 mi Erienmeyet 4 You wil take 25 s uve the vernier labquest and ple electrode to record pat data. First. alibrate the vernier labquest with a ene pant alibration using the pHa4 standard ntrate the malonic acid solution with 025M NaoH tothn two equivalence points he equivalence point 8. Repeat experiment at ry in order to pet Calculate the pH dentify the major species and calculate the pH before Naot is added. he fast equivalent point 2 identify the major species and calculate the pH ldentify the major species and calculate the pH at the second equivalem poim. 4. Graph the titration data: volume to NaoH usedvs pH and label the first and second an the graph 15mL of used In titrnyinm

Explanation / Answer

Malonic acid titration with NaOH

lets say we started with 0.2 g of malonic acid

concentration of malonic acid solution = 0.2 g/104.06 g/mol x 0.1 L = 0.02 M

25 ml aliquit titrated

pH

a) before adding NaOH

let x amount of acid has dissociated

H2A + H2O <==> HA- + H3O+

Ka1 = [HA-][H3O+]/[H2A]

1.5 x 10^-3 = x^2/0.02

x = [H3O+] = 5.48 x 10^-3 M

pH = -log[H3O+] = 2.26

b) pH at first half equivalence point

[H2A] remaining = [HA-] fomed

pH = pKa1

pKa = -log[Ka] = 2.82

c) at first equivalence point

pH = 1/2(pKa1 + pKa2)

     = 1/2(2.82 + 5.70) = 4.26

d) at second half equivalence point

[HA-] left = [A^2-] formed

pH = pKa2

pKa2 = -log[Ka2] = 5.70

e) at the second equivalence point

volume of NaOH added = 0.02 M x 25 ml/0.25 M = 2 ml

[A^2-] formed = 0.02 m x 25 ml/27 ml = 0.02 M

A^2- + H2O <==> HA- + OH-

let x amount has hydrolyzed

Kb1 = Kw/Ka2 = [HA-][OH-]/[A^2-]

1 x 10^-14/2 x 10^-6 = x^2/0.02

x = [OH-] = 1 x 10^-5 M

pOH = -log[OH-] = 5

pH = 14 - pH = 9

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