Match the following terms with their definitions:
2 answers:
Explanation:
A mixture is defined as a substance in which two or more components are mixed together. Components in a mixture do not chemically combine together.
For example, sugar in water.
When solute particles completely and evenly mixed in a solvent then it is known as a homogeneous mixture. A homogeneous mixture is also known as a solution. For example, salt in water.
When solute particles are not evenly mixed in a solvent then it is known as a heterogeneous solution. For example, sand in water.
A compound is a substance in which two or more atoms are chemically combined together in a fixed ratio by mass. For example, is a compound.
Thus, we can conclude that given terms with their definitions are matched as follows.
1. Matter that is physically combined; has two forms of itself (homogeneous and heterogeneous) - Mixture.
2. A substance that that has uniform composition of two or more substances that cannot be seen - homogeneous mixture.
3. A substance that that has non-uniform composition of two or more substances that can be seen - heterogeneous mixture.
4. Substance composed of two or more elements bonded in a certain ratio by mass - Compound.
5. An example of a homogeneous mixture in a liquid form - solution.
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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.
Hello!
First, we can assume the density of water to be 1 g/mL, so the volume of water would be 725 mL. The density of Acetic Acid (Pure) is 1,05 g/mL so 8 grams would represent 7,61 mL
Now we can apply the following conversion factor to calculate the molarity of the solution, using the molar mass of Acetic Acid:
![[CH_3CO_2H]=\frac{8 g_{CH_3CO_2H} }{725 mL + 7,61 mL}* \frac{1000 mL}{1 L}* \frac{1 mol_{CH_3CO_2H} }{60,05 g_{CH_3CO_2H} }](https://tex.z-dn.net/?f=%5BCH_3CO_2H%5D%3D%5Cfrac%7B8%20g_%7BCH_3CO_2H%7D%20%20%7D%7B725%20mL%20%2B%207%2C61%20mL%7D%2A%20%5Cfrac%7B1000%20mL%7D%7B1%20L%7D%2A%20%5Cfrac%7B1%20mol_%7BCH_3CO_2H%7D%20%7D%7B60%2C05%20g_%7BCH_3CO_2H%7D%20%7D)
![[CH_3CO_2H]=0,1818 M](https://tex.z-dn.net/?f=%5BCH_3CO_2H%5D%3D0%2C1818%20M)
So, the concentration of acetic acid would be 0,1818 M
Have a nice day!
First, we can assume the density of water to be 1 g/mL, so the volume of water would be 725 mL. The density of Acetic Acid (Pure) is 1,05 g/mL so 8 grams would represent 7,61 mL
Now we can apply the following conversion factor to calculate the molarity of the solution, using the molar mass of Acetic Acid:
So, the concentration of acetic acid would be 0,1818 M
Have a nice day!