Molecular weight of PbCl2 = 278 g/mole Mass of beaker + PbCl2 = 62.19 g Mass of beaker = 57.77 g Mass of PbCl2 per 100 mL = g Mo
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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
6,0 = 6,4 + log₁₀
0,3981 = <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× = <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.