Bromine, a liquid at room temperature, has a boiling point of 58°C and a melting point of –7.2°C. Bromine can be classified as a
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Use stoichiometric ratio to determine which is in excess and which is limiting
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.