Answer:
0,542 g of Na₂HPO₄ and 0,741 g of NaH₂PO₄.
0,856 g of Tris-HCl and 0,553 g of Tris-base
Explanation:
It is possible to use Henderson–Hasselbalch equation to estimate pH in a buffer solution:
pH = pka + log₁₀ [tex]\frac{A^{-} }{HA}[/tex]
Where A⁻ is conjugate base and HA is conjugate acid
The equilibrium of phosphate buffer is:
H₂PO₄⁻ ⇄ HPO4²⁻ + H⁺ Kₐ₂ = 6,20x10⁻⁸; pka=7,21
Thus, Henderson–Hasselbalch equation for phosphate buffer is:
pH = 7,21 + log₁₀ [tex]\frac{HPO4^{2-} }{H2PO4^{-} }[/tex]
If desire pH is 7,0 you will obtain:
0,617 = [tex]\frac{HPO4^{2-} }{H2PO4^{-} }[/tex] (1)
Then, if desire concentration of buffers is 0,10 M:
0,10 M = [HPO₄²⁻] + [H₂PO₄⁻] (2)
Replacing (1) in (2) you will obtain:
[H₂PO₄⁻] = 0,0618 M
And with this value:
[HPO₄²⁻] = 0,0382 M
As desire volume is 100mL -0,1L- the weight of both Na₂HPO₄ and NaH₂PO₄ is:
Na₂HPO₄ = 0,1 L× [tex]\frac{0,0382mol}{1L}[/tex]× [tex]\frac{141,96g}{1mol}[/tex] = 0,542 g of Na₂HPO₄
NaH₂PO₄ = 0,1 L× [tex]\frac{0,0618mol}{1L}[/tex]× [tex]\frac{119,96g}{1mol}[/tex] = 0,741 g of NaH₂PO₄
For tris buffer the equilibrium is:
Tris-base + H⁺ ⇄ Tris-H⁺ pka = 8,075
Henderson–Hasselbalch equation for tris buffer is:
pH = 8,075 + log₁₀ [tex]\frac{Tris-base }{Tris-H^{+} }[/tex]
If desire pH is 8,0 you will obtain:
0,841 = [tex]\frac{Tris-base }{TrisH^{+} }[/tex] (3)
Then, if desire concentration of buffers is 0,10 M:
0,10 M = [Tris-base] + [Tris-H⁺] (4)
Replacing (3) in (4) you will obtain:
[Tris-HCl] = 0,0543 M
[Tris-base] = 0,0457 M
As desire volume is 100mL -0,1L- the weight of both Tris-base and Tris-HCl is:
Tris-base = 0,1 L× [tex]\frac{0,0457mol}{1L}[/tex]× [tex]\frac{121,1g}{1mol}[/tex] = 0,553 g of Tris-base
Tris-HCl = 0,1 L× [tex]\frac{0,0543mol}{1L}[/tex]× [tex]\frac{157,6g}{1mol}[/tex] = 0,856 g of Tris-HCl
I hope it helps!
