You are given an unknown diprotic acid. Titraiton of an aqueous solution of this

Question

You are given an unknown diprotic acid. Titraiton of an aqueous solution of this acid with a 2.0 M sodium hydroxide solution is conducted. Using the given experimental data, determine the molar mass of this acid and identify it from the list of possibilities.

Part A

Mass of unkown acid: 1.33 grams

Volume of base used: 9.90 mL

Part B

Assmming you used the given amount of acid in a n initial volume of 25 mL, compete the following table.

Acid in 50 mL aq. soln. (before any base added)

*I have figure the first part out using the list of acids. The answer to Part A is malic acid (C4H6O5), Molecular weight: 134.09. Now, i just do not know how to finish part B, C, D, E.

Part C

Sketch a diagram of what the titration curve should look like for the above titration. Make sure you plot has approtriate labels.

Part D

Your acid undergoes a decarboxylation reaciton (loss of CO2) to give a monoprotice acided molecule. Kinetic studies indicate that the rat of this reaciton (in ploar nonaqueos solvents) occurs at a rate of 0.05 M/s at 50 degrees Celcius when the initila concentaration of the acid is 0.05 M.

               a) Write the balanced chemical equation for this reaciton.  

               b) Write the rate expression for this reaciton based on alll reactants and products.

Part E

Now that you know a lot about your acid compound. Look up its pKa values adn explain how you could prepare a buffer solution at a pH of 5.0 using your acid.

    

[H30], (M) [OH-], (M) pH        pOH     

Acid in 50 mL aq. soln. (before any base added)

After addition of 1 molar equivilant of NaOH After addition of 45 mL of NaOH (aq)

Explanation / Answer

PART-A

ACID (diprotic) Base (NaOH)

mass (m) = 1.33g M2 (molarity) = 2.0M

Mol.wt=? V2 (volume)= 9.9 mL   

n1(no.of moles)= 1 n2 (no.of moles)= 2

as diprotic acid reacts with mono basic NaOH , one mole of the acid reacts with two moles of the base.

according to law of equivalence, M1V1/n1 = M2V2/n2

[wt./g.mol.wt * 1000/V(in mL] * V1/ n1 = 2*9.9/2

   1.33/g.mol.wt * 1000/V * V1/n1 =9.9

[1.33 * 1000] / x = 9.9,

gram molecular weight of acid, x= 134.34g/mol

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