How can you separate the components soil and water
1 answer:
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Answer:
0.11 M
Explanation:
The computation of molarity of the CH3COOH solution is shown below:-
Where,
= Molarity of LioH = 0.0175
= Volume of LioH = 18.5 ml
= Volume of CH3COOH = 29.4 ml
Now, we will put the values into the formula
Which gives result
= 0.011 M
Therefore for computing the molarity of the CH3COOH solution we simply applied the above formula.
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
Where A⁻ is conjugate base and HA is conjugate acid
The equilibrium of phosphate buffer is:
H₂PO₄⁻ ⇄ HPO4²⁻ + H⁺ Kₐ₂ = 6,20x10⁻⁸; <em>pka=7,21</em>
Thus, Henderson–Hasselbalch equation for phosphate buffer is:
pH = 7,21 + log₁₀
If desire pH is 7,0 you will obtain:
<em>0,617 = </em><em>(1)</em>
Then, if desire concentration of buffers is 0,10 M:
0,10 M = [HPO₄²⁻] + [H₂PO₄⁻] <em>(2)</em>
Replacing (1) in (2) you will obtain:
<em>[H₂PO₄⁻] = 0,0618 M</em>
And with this value:
<em>[HPO₄²⁻] = 0,0382 M</em>
As desire volume is 100mL -0,1L- the weight of both Na₂HPO₄ and NaH₂PO₄ is:
Na₂HPO₄ = 0,1 L× ×
= 0,542 g of Na₂HPO₄
NaH₂PO₄ = 0,1 L× ×
= 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
If desire pH is 8,0 you will obtain:
<em>0,841 = </em><em>(3)</em>
Then, if desire concentration of buffers is 0,10 M:
0,10 M = [Tris-base] + [Tris-H⁺] <em>(4)</em>
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× ×
= 0,553 g of Tris-base
Tris-HCl = 0,1 L× ×
= 0,856 g of Tris-HCl
I hope it helps!