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Zarrin [17]
3 years ago
12

Phosphoric acid is a triprotic acid with the following pKa values:pKa1=2.148, pKa2=7.198, pKa3=12.375You wish to prepare 1.000 L

of a 0.0500 M phosphate buffer at pH 7.540. To do this, you choose to mix the two salt forms involved in the second ionization, NaH2PO4 and Na2HPO4, in a 1.000 L volumetric flask and add water to the mark. What mass of each salt will you add to the mixture?Mass NaH2PO4Mass Na2HPO4What other combination of phosphoric acid and/or its salts could be mixed to prepare this buffer? (Check all that apply).H3PO4 and NaH2PO4H3PO4 and Na2HPO4H3PO4 and Na3HPO4NaH2PO4 and NA3PO4Na2HPO4 and NA3PO4PLEASE SOLVE AND EXPLAIN
Chemistry
1 answer:
laila [671]3 years ago
8 0

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] = 10^{0.342}

[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.

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Explanation:

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a) The concentration of the drug in the bottle is 294 mg/ 30.0 ml = 9.8 mg/ml

b) The drug has to be administrated at a dose of 0.0210 mg/ kg body mass. Then, the total mass of drug that there should be in the injection for a person of 70 kg will be:

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The volume of solution that contains that mass of drug can be calculated using the value of the concentration calculated in a)

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To prepare the injection, you should take 0.15 ml of the concentrated drug solution and (2.0 ml - 0.15 ml) 1.85 ml saline

c) In the injection there is a concentration of (1.47 mg / 2.0 ml) 0.735 mg/ml.

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you can now deduct which one is greater or smaller and by how much.

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ii) 0.6= m/( 2*1 + 16)

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2. Avogadro's Number = 6.02×10²³

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bi) 0.25 mol of HCl= (6.02×10²³) * 0.25

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