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Como preparar soluciones
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How to make a solution:

The following "how to make a solution" is from CAROLINA

Remember: Always follow appropriate lab safety procedures when mixing and storing chemicals.
When a substance, called a solute, is dissolved in another substance, called the solvent, a solution is formed. A solution is a uniform distribution of solute in solvent. For example, vinegar is a solution of acetic acid, the solute, in water, the solvent. The amount of solute in a solvent is important and can be expressed in several different ways. Some common units of concentration will be discussed in this manual.

1.
Molar solutions

Molarity (M) means the number of moles of solute per liter of solution. To prepare a 1 M solution, slowly add the molecular weight (x g/mol) of compound to 500-mL distilled or deionized water in a 1000-mL volumetric flask half filled with distilled or deionized water. Allow the compound to dissolve completely, swirling the flask gently if necessary. Once the solute is completely dissolved and the solution is at room temperature, dilute to the mark with water. Invert the flask several times to mix.

To make a 1 M solution of sodium hydroxide (40g/mol), slowly add 40 g sodium hydroxide to 500 mL distilled or deionized water in a 1000-mL volumetric flask. When all the solid is dissolved and the solution is at room temperature, dilute to the mark and invert the flask several times to mix.To make a 1 M solution of acetic acid, dissolve 60.05 g acetic acid in 500 mL distilled or deionized water in a 1000-mL volumetric flask. Since acetic acid is a liquid, it may also be measured by volume. Divide the mass of the acid by its density (1.049 g/mL) to determine the volume (57.24 mL). Use either 60.05 g or 57.24 mL acetic acid to make the solution. Swirl the flask gently to mix the solution. When the solution is at room temperature, dilute to the mark and invert the flask several times to mix. Always add acid to water

2.
Percent solutions

a.
Mass percent means the number of grams of solute per 100 g of solution. For example, 10 g sodium chloride in 90 g water is a 10% by mass solution.

mass percent = mass of solute/mass of solution= 10 g / (10 g + 90 g) x 100%= 10%

b.
Volume percent means the number of milliliters of solute per 100 mL of solution. The volume percent of a solution cannot be calculated directly from the volumes of its components because the final volume may not equal the sum of the components’ volumes. To prepare volume percent solutions, first determine the final volume and concentration of the desired solution and then determine the amount of solute. Dilute the solute in sufficient solvent to produce the final volume of desired solution. For example, to prepare 100 mL of a 10% by volume solution of acetic acid, dilute 10 mL acetic acid with distilled or deionized water to make 100 mL of solution.
Note: Solutions of concentrated reagents, such as 37% hydrochloric acid and 85% phosphoric acid, are percent solutions by mass. In general, percent solutions are by mass.

3.
Dilutions

When preparing a dilution, decide the volume and molar concentration of the resulting solution that you require. Use the following equation to determine how much of the concentrated reagent is needed to prepare the diluted solution,

M reagent x V reagent = M dilution x V dilution

where M is molarity and V is volume.

Slowly add the calculated volume of concentrated reagent to a proper-sized volumetric flask half filled with distilled or deionized water and swirl the flask to mix. Once the solution is at room temperature, dilute to the mark with water and invert the flask several times to mix.

For example, what volume of 10 M acetic acid is required to prepare 1.0 L of 0.50 M acetic acid?

10 M x V reagent = 0.50 M x 1.0 L V reagent = 0.050 L = 50 mL

A volume of 50 mL of 10 M acetic acid is required to prepare 1.0 L of 0.50 M acetic acid.

4.
Special cases

Often it is necessary to prepare solutions from chemicals that are less than 100% pure. To prepare solutions from these impure chemicals, first decide the volume and molarity of the required solution. Multiply the solution’s volume by its molarity. The product (n) is the number of moles of pure chemical needed to produce that solution.

M pure x V pure = n pure

Because the percent purity of chemicals sold commercially is measured by mass, first calculate the mass of the pure chemical needed to make the solution. Multiply the number of moles of pure chemical times the gram formula weight of the chemical.

mass of pure chemical = n pure x gram formula weight

The mass of the impure chemical times the percent purity in decimal form equals the mass of the pure chemical. Divide the mass of pure chemical by the percent purity to yield the mass of the impure chemical.
mass of pure chemical = mass of impure chemical x percent purity mass of impure chemical = mass of pure chemical / percent purity

The above equations may be combined into one equation.

mass of impure chemical = M pure x V pure x gram formula weight / percent purity

Use this equation if the chemical in question is a solid.

For example, what mass of potassium hydroxide that is 85.9% pure is needed to prepare 1.0 L of a 0.25 M solution of potassium hydroxide? The gram formula weight of potassium hydroxide is 56.11 g/mol.

mass of impure chemical = M pure x V pure x gram formula weight/percent purity
= 0.25 M × 1.0 L × 56.11 g/mol ÷ 0.859= 16 g

If the chemical in question is a liquid, then one more calculation is required. Divide the mass of the impure chemical by its density to yield the volume of chemical.

volume of impure chemical in mL = mass of impure chemical in g/density of impure chemical in grams per milliliter
Again, combine the previous equations.

volume of impure chemical = M pure x V pure x gram formula weight / (percent purity x density)
For example, what volume of hydrochloric acid that is 37.1% pure is needed to prepare 1.0 L of a 0.10 M solution of hydrochloric acid? The gram formula weight of hydrochloric acid is 36.46 g/mol, and the density of 37.1% hydrochloric acid is 1.2 g/mL.

volume of impure chemical = M pure x V pure x gram formula weight / (percent purity x density)
= 0.10 M × 1.0 L × 36.46 g/mol ÷ (0.371 x 1.2 g/mL)= 8.2 mL

5.
Normal solutions

Another concentration term sometimes used is normality. Normality (N) means the number of equivalents of solute per liter of solution. An equivalent is defined separately for acid-base and reduction-oxidation (redox) chemistry.
In acid-base chemistry, an equivalent is the mass of chemical that donates or accepts one mole of protons. For example, sulfuric acid is a diprotic acid. One-half mole of sulfuric acid therefore provides one mole of protons. To prepare a 1 N solution of sulfuric acid, slowly add 0.5 gram formula weight of sulfuric acid to a clean 1-L volumetric flask and fill to the mark with distilled or deionized water.

In redox chemistry, an equivalent is the mass of chemical that donates or accepts one mole of electrons. Determine the number of electrons a chemical donates or accepts from its half-reaction. For example, one mole of aluminum (III) reacts with 3 moles of electrons to give one mole of aluminum metal.

Al 3+ + 3e – Al

To prepare a 1 N solution of aluminum (III), slowly add 0.333 g formula weight of an aluminum (III) compound to a clean 1-L volumetric flask and fill to the mark with distilled or deionized water.