The following questions are based on a triprotic acid (H 3 A) with pK 1 = 2.20,

Question

The following questions are based on a triprotic acid (H3A) with pK1 = 2.20, pK2 = 9.11 and pK3 = 10.07. The disodium salt of the acid was dissolved to make 60.00 mL of a 0.040 M soluion and was then titrated with 0.100 M HCl.

a) Write the balanced chemical equation for the reaction of the disodium salt with HCl to reach the final equivalence point.

b) How many equivalence points will be observed during the titration?

c) Calulate the pH of the solution before any HCl is added (Va = 0 mL).

d) What was the pH of the solution when Va = 36 mL?

Explanation / Answer

Solution:

The computation are expressed below,

a) C10H14N2Na2O8 + 2HCl = C10H16N2O8 + 2NaCl

b) Total of three equivalence points are observed as three midpoints will be observed in the plot from titration process of disodium salt with HCl

c) by using Henderson-Hasselbalch equation the pH is computed,

pH = pKa + ([Conj.base]/[Conj.acid])

for pK1 = 2.20 gives pH = 3.59

for pK2 = 9.11 gives pH = 11.51

for pK3 = 10.07 gives pH = 12.46

Hence our titration gives increase of plot from acidic to basic condition as expectd from titration.

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