i know the answer to part a) and b) but need help in the other three! please hel

Question

i know the answer to part a) and b) but need help in the other three! please help!

6. Polpyprotic Titration Curves The amino acid, alanine (CHTNO), is diprotic with a pKal 2.34 and pKa2 9.87. In this problem, 89.1 mg of this amino acid is titrated with 0.1 M NaOH titrant. a) Where is equivalence point #1? b) What is the equation used to find the theoretical pH at equivalence point #1? c) What should the pH at point A be and why? d) What should the pH at point C be and why? e) How can you calculate the molar mass of this compound from the experimental data? Show the calculations 15 20 25 30 Titrant volume (mL)

Explanation / Answer

c) A is the half equivalence point for 1st equivalence point and at pH at half equivalence point is equal to pKa.

So, pH at point A = pKa1 = 2.34

Using the formula, pH = pKa + log [salt]/[acid]

At half equivalence point, [salt] = [acid], so pH = pKa

d) Similarly, C is the half equivalence point for 2nd equivalence point and at pH at half equivalence point is equal to pKa.

So, pH at point C = pKa2 = 9.87

e) Concentration of NaOH used = 0.1 M

volume of NaOH used (till 2nd equivalence point) = 20 ml = 0.020 L

Moles of NaOH used = 0.1*0.02 = 0.002 mol

Amino acid is diprotic acid, so 2 moles of NaOH are used for 1 mol of amino acid.

So, moles of amino acid = 0.002/2 = 0.001 mol

mass of amino acid used = 89.1 mg = 0.0891 g

Molar mass of amino acid = mass/moles = 0.0891/0.001 = 89.1 g/mol

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