Describe how you would prepare 100 mL of a 0.050 M sodium citrate tribasic buffer to a pH of 6. This method is accomplished by preparing a solution of a weak species base and then adding the appropriate strong species 0.5 M HCl to get to the desired pH. What amount of mass or volumes should you add?

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02-10-2019
Chemistry

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Describe how you would prepare 100 mL of a 0.050 M sodium citrate tribasic buffer to a pH of 6. This method is accomplished by preparing a solution of a weak species base and then adding the appropriate strong species 0.5 M HCl to get to the desired pH. What amount of mass or volumes should you add?

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jufsanabriasa jufsanabriasa · Answer 1 · 2019-10-08T22:27:03+02:00

Answer:

To prepare 100 mL of a 0,050M sodium citrate tribasic buffer to a pH of 6 you need to add 7,16 mL of 0,5M HCl, 1,4705 g of sodium citrate tribasic dihydrate and complete 100 mL with water.

Explanation:

The acid equilibrium of sodium citrate tribasic buffer is:

citrate dibasic⁻² ⇄ citrate tribasic⁻³ + H⁺ pka = 6,4

Using Henderson-Hasselbalch formula:

pH = pka + log₁₀ [tex]\frac{[A^{-}]}{[HA]}[/tex]

6,0 = 6,4 + log₁₀ [tex]\frac{[CitrateTribasic]}{[CitrateDibasic]}[/tex]

0,3981 = [tex]\frac{[CitrateTribasic]}{[CitrateDibasic]}[/tex] (1)

You need to add 0,1L× 0,050M = 0,0050moles of sodium citrate:

0,0050 moles = Citrate tribasic + Citrate dibasic (2)

Replacing (2) in (1)

Citrate dibasic: 3,58x10⁻³ moles

Thus,

Citrate tribasic: 1,42x10⁻³ moles

Citrate tribasics reacts with HCl thus:

Citrate tribasic⁻³ + HCl → Citrate dibasic⁻² + Cl⁻

Thus, you need to add 5x10⁻³moles of sodium citrate tribasic and 3,58x10⁻³ moles of HCl:

5x10⁻³moles of sodium citrate tribasic×[tex]\frac{294,1 g}{1mole}[/tex] = 1,4705 g of sodium citrate tribasic dihydrate -commercial reactant-

3,58x10⁻³ moles of HCl÷ 0,5 M = 7,16x10⁻³ L ≡ 7,16 mL of 0,5M HCl

Thus, to prepare 100 mL of a 0,050M sodium citrate tribasic buffer to a pH of 6 you need to add 7,16 mL of 0,5M HCl, 1,4705 g of sodium citrate tribasic dihydrate and complete 100 mL with water.