Answer:
Explanation:
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In this case, since the mole fraction relates the moles of solute and the moles of solution, and for sucrose it is:
Whereas we need to compute the moles of both sucrose and water based on their molar masses (342.3 and 18.02 respectively) as shown below:
Thus, the mole fraction sucrose turns out:
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Answer:
NaH₂PO₄ = 1.876 g
Na₂HPO₄ = 4.879 g
Combinations: H₃PO₄ and Na₂HPO₄; H₃PO₄ and Na₃HPO₄
Explanation:
To have a buffer at 7.540, the acid must be in it second ionization, because the buffer capacity is pKa ± 1. So, we must use pKa2 = 7.198
The relation bewteen the acid and its conjugated base (ion), is given by the Henderson–Hasselbalch equation:
pH = pKa + log[A⁻]/[HA], where [A⁻] is the concentration of the conjugated base, and [HA] the concentration of the acid. Then:
7.540 = 7.198 + log[A⁻]/[HA]
log[A⁻]/[HA] = 0.342
[A⁻]/[HA] =
[A⁻]/[HA] = 2.198
[A⁻] = 2.198*[HA]
The concentration of the acid and it's conjugated base must be equal to the concentration of the buffer 0.0500 M, so:
[A⁻] + [HA] = 0.0500
2.198*[HA] + [HA] = 0.0500
3.198*[HA] = 0.0500
[HA] = 0.01563 M
[A⁻] = 0.0500 - 0.01563
[A⁻] = 0.03436 M
The mix reaction is
NaH₂PO₄ + Na₂HPO₄ → HPO₄⁻² + 3Na + H₂PO₄⁻
The second ionization is:
H₂PO₄⁻ ⇄ HPO₄⁻² + H⁺
So, H₂PO₄⁻ is the acid form, and its concentration is the same as NaH₂PO₄, and HPO₄⁻² is the conjugated base, and its concentration is the same as Na₂HPO₄ (stoichiometry is 1:1 for both).
So, the number of moles of these salts are:
NaH₂PO₄ = 0.01563 M * 1.000 L = 0.01563 mol
Na₂HPO₄ = 0.03436 M* 1.000 L = 0.03436 mol
The molar masses are, Na: 23 g/mol, H: 1 g/mol, P: 31 g/mol, and O = 16 g/mol, so:
NaH₂PO₄ = 23 + 2*1 + 31 + 4*16 = 120 g/mol
Na₂HPO₄ = 2*23 + 1 + 31 + 4*16 = 142 g/mol
The mass is the number of moles multiplied by the molar mass, so:
NaH₂PO₄ = 0.01563 mol * 120 g/mol = 1.876 g
Na₂HPO₄ = 0.03436 mol * 142 g/mol = 4.879 g
To prepare this buffer, it's necessary to have in solution the species H₂PO₄⁻ and HPO₄⁻², so it can be prepared for mixing the combination of:
H₃PO₄ and Na₂HPO₄ (the acid is triprotic so, it will form the H₂PO₄⁻ , and the salt Na₂HPO₄ will dissociate in Na⁺ and HPO₄²⁻);
H₃PO₄ and Na₃HPO₄ (same reason).
The other combinations will not form the species required.
The chemical reaction in which number of atoms of each element present in the reactant side is equal to the number of atoms of that element in product side, such reactions are said to be a balanced chemical reaction.
The chemical symbol for sodium is .
The chemical symbol for fluorine gas is .
The chemical symbol for sodium fluoride is
The sodium fluoride is prepared from the reaction between sodium metal and fluorine gas can be written as:
The above reaction is not balanced as the number of fluorine atoms are not same on reactant and product side. So, in order to balance the reaction we will multiply with 2 on reactant side and
with 2 on product side. Thus, the balanced reaction will be:
Thus, the balanced chemical equation is .