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vagabundo [1.1K]
4 years ago
9

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 p

reparing 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?
Chemistry
1 answer:
Vinvika [58]4 years ago
6 0

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⁺ <em>pka = 6,4</em>

Using Henderson-Hasselbalch formula:

pH = pka + log₁₀ \frac{[A^{-}]}{[HA]}

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

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

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

0,0050 moles = Citrate tribasic + Citrate dibasic <em>(2) </em>

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×\frac{294,1 g}{1mole} = <em>1,4705 g of sodium citrate tribasic dihydrate </em><em>-commercial reactant-</em>

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

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.

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