Calculations Associated with Solutions

Calculations Associated With Solutions

PERCENT SOLUTIONS

Solutions may be expressed in other concentrations besides molarity, molality, or normality. In the laboratory, solutions are sometimes expressed in terms of relative percent concentration of solute to solution. There are three types of percent concentrations that may be used in the laboratory. In all three types, the total volume of the solvent is based on a quantity of 100 mL or gram, not 1.00 L, as in molarity or normality concentrations.

Percent Weight/Weight

A percent weight/weight (%^w/w) solution is calculated using the following formula:

%W/W=gramofsolute100.0gofsolutionORgram of solute per100.0g of solution

In this type of solution, the amounts of solute and solvents are weighed individually using a balance. It is important to note that as the total is based on 100.0 g of solution, the amount of solvent that must be weighed is determined by subtracting the quantity of solute needed from 100.0. A solution consists of solute + solvent. Once the amount of solvent is determined, the solvent and solute are combined and mixed together in a flask or beaker. In the clinical laboratory, deionized water is the most frequently used solvent. The %^w/w solutions are the most accurate because unlike %^w/v, their concentrations do not fluctuate with temperature. However, the %^w/w solutions are not commonly prepared in the clinical laboratory.

Example 6–1

How many grams of NaOH are needed to make a 30.0%^w/w solution using deionized water as the solvent?

To solve this problem, remember the basic formula for %^w/w, and substitute the numbers that are known.

To make this solution, 30.0 g of NaOH would be dissolved into 70.0 g of H₂O in a beaker or flask.

Additional Examples

How many grams of HCl are necessary to make a 10% HCl^w/w solution using deionized water as the solvent?

Using the formula for %^w/w, 10 g of HCl would be dissolved in 90 g of H₂O in a beaker or flask.

How much solvent would be needed to make a 20%^w/w solution of CuSO₄?

To make this solution, 80 g of solvent would be needed. This question asked for the amount of solvent, not the amount of solute (CuSO₄).

Percent Weight/Volume

A percent weight/volume (%^w/v) solution is calculated by the following formula:

%W/W=gofsolute100mLofsolutionORgofsoluteper100.0mLofsolution

The %^w/v solution is the most frequently used percent solution in the clinical laboratory. For this solution, the amount of solute is weighed on a balance and then placed into a 100-mL volumetric flask in which there is a small amount of solvent to dissolve the solute. Once the solute is dissolved, the remaining solvent (which in most cases in the clinical laboratory is deionized water) is then added to the volumetric flask to the calibrated mark. The Latin term quantis satis, or quantity sufficient (qs) is often used in the laboratory to describe the addition of the solvent to the calibrated mark. For example, you may be instructed to add a determined amount of solute to a volumetric flask and then qs it to the calibration mark with the appropriate solvent.

Example 6–2

What is the %^w/v of a solution that has 25.0 g of NaCl dissolved into a total volume of 100 mL deionized water?

To solve this problem, use the formula for %^w/v:

Therefore, a solution with 25.0 g of NaCl dissolved into 100 mL of water has a %^w/v concentration of 25.0%.

Additional Examples

How many grams of NaOH would be needed to make a 40%^w/v solution using deionized water as the solvent?

To make a 40%^w/v solution, 40.0 g of NaOH would be dissolved into 100 mL of water.

How many grams of CaCl₂ would be needed to make a 10%^w/v solution using deionized water as the solvent?

To make a 10%^w/v solution, 10 g of CaCl₂ would be dissolved into 100 mL of water.

Percent Volume/Volume

A percent volume/volume (%^v/v) solution uses volume, or liquid, measurements for both the solute and the solvent.

A %^v/v solution is calculated by the following formula:

%V/V=mLofsolute100mLofsolutionORmilliliterofsoluteper100mLofsolution

Percent ^v/v is similar to %^w/w in that the total volume of the solution is 100 mL.

Therefore, the amount of solvent is determined by the following formula:

100mLtotalvolume−amountofsolute=amountofsolvent

Example 6–3

How many milliliters of ethanol (EtOH) are needed to make a 75.0%^v/v solution using deionized water as the solvent?

This problem is solved in the same manner as solving for %^w/w and %^w/v. Using the formula for %^v/v and substituting in the appropriate given numbers, the following equation is derived:

How many milliliters of water are necessary for this solution?

The amount of water needed is determined by the following formula:

Amount of solvent=100−amount of soluteX=100−75.0X=25.0

Therefore, 75.0 mL of EtOH would be added to 25.0 mL of deionized water to produce a solution of 75.0%^v/v EtOH.

Additional Examples

How many milliliters of HCl are needed to make a 10%^v/v solution using deionized water as the solvent?

To make a 10%^v/v solution, 10 mL of HCl are added to 90 mL of deionized water.

How many milliliters of H₂SO₄ are needed to make a 2.5%^v/v solution using deionized water as the solvent?

To make a 2.5%^v/v solution, 2.5 mL of H₂SO₄ are added to 97.5 mL of deionized water.

Conversion of Percentage Calculations to Molarity

In the laboratory, sometimes a chemical solution may be labeled in terms of relative percentage but you may need to know the molarity of that solution. By using both molarity and percentage formulas, you can interchange between the two types of concentrations.

Example 6–4

Suppose you found a bottle labeled 0.85%^w/v NaCl in your laboratory. What is the molarity of this solution?

To solve this problem, first use the percent formula to determine the concentration of NaCl in the solution. A %^w/v solution is calculated by the number of g of solute dissolved in 100 mL of solvent. As we have a 0.85%^w/v solution, we know that there are 0.85 g of NaCl dissolved in 100 mL of water. Because molarity is based on the concentration in 1.00 L of solute, we need to determine the quantity of NaCl present in 1.00 L. We can use a simple ratio formula to determine this:

Using algebra and crossmultiplying we derive the following:

Next, use the molarity formula to determine the molarity of this solution. We know that there are 8.50 g of NaCl present per liter of solvent in this solution. The molecular weight of NaCl is 58.44. Next, substitute all of the known values to determine the molarity of this solution:

Therefore, an 0.85%^w/v NaCl solution has a molarity of 0.14 mol/L.

Additional Examples

What is the molarity of a 40%^w/v NaOH solution made with deionized water as the solvent?

To solve, first determine how many grams of NaOH is in the 40%^w/v solution. You know that a 40%^w/v solution of NaOH contains 40 g of NaOH in 100 mL of water. In Example 6–4 the decimal point was moved one place to the right to convert 0.85 g/100 mL to 8.5 g/1000 mL. Therefore, by moving the decimal point one place to the right in this example, 40 g NaOH in 100 mL of water is the same as 400 g of NaOH in 1000 mL of water. Next, determine the gram molecular weight of NaOH by determining the sum of the gram atomic weights of Na, O, and H. The gram molecular weight of NaOH is 40.00. Next, substitute the values obtained so far in this problem into the molarity formula:

Therefore, a 40.0%^w/v NaOH solution has a molarity of 10.0 mol/L.

What is the molarity of a 10.0%^w/v CaCl₂ solution?

To solve, 10.0%^w/v is equal to 10.0 g/100 mL or 100 g/1000 mL. The gram molecular weight of CaCl₂ is 110.98. Therefore,

Therefore, a 10%^w/v CaCl₂ solution has a molarity of 0.90 mol/L.

Example 6–5

What would be the normality of a solution of 15.0%^w/v H₂SO₄?

To solve this problem, remember that

To determine the amount of g in 1.00 L:

Crossmultiplying the equations results in the following:

Next, use the normality formula to determine the normality of the solution.

Therefore, a 10.0%^w/v H₂SO₄ solution has a normality of 3.06 Eq/L.

Additional Examples

What is the normality of a 12.5%^w/v NaOH solution?

To solve, use the normality formula after determining the number of grams per liter and the gram equivalent weight of NaOH.

Therefore, a 12.5%^w/v solution of NaOH has a normality of 3.12 Eq/L.

What is the normality of a 10%^w/v H₂CO₃ solution?

To solve, use the normality formula after determining the number of grams per liter and the gram equivalent weight of H₂CO₃.

Therefore, a 10%^w/v H₂CO₃ solution has a normality of 3.22 Eq/L.

ANHYDROUS VERSUS HYDRATED SOLUTIONS

Another type of solution may consist of hydrated chemical salts dissolved in a solvent. As the difference between the hydrous and the anhydrous form of a chemical is the amount of water molecules present in the hydrated form, it may be necessary to be able to interchange between the two. Sometimes, in the laboratory, only one of the forms of a chemical is available. You must be able to interchange the two forms to arrive at the form that you need. The basic formula for interchanging between hydrated and anhydrous chemicals is as follows:

anhydrouschemicalmolecularweighthydratedchemicalmolecularweight=gramofanhydrouschemicalgramofhydratedchemical

Notice that it is a ratio and proportion calculation. Ratio and proportion is used whenever the same concentration of a solution but different quantities of the solution is necessary. In anhydrous as opposed to hydrated forms of a chemical, the ratio and proportion are used to ensure that the same concentration of the chemical is used regardless of its water content.

Example 6–6

Suppose you needed to make a 12.0%^w/v solution of CuSO₄, but only CuSO₄ · H₂O was available. How much of the CuSO₄ · H₂O should you use? The gram molecular weight of CuSO₄ is 159.61, whereas the gram molecular weight of CuSO₄ · H₂O is 177.63.

The first step is to determine the amount of grams in a 12.0%^w/v solution. A 12%^w/v solution contains 12.0 g in 100 mL of solvent. Using the formula for interchanging hydrated and anhydrous chemicals, the following equation is derived:

By crossmultiplying, the following equation is derived:

Therefore, 13.4 g of CuSO₄ · H₂O qs to 100 mL of solvent will result in a 12.0 %^w/v solution of CuSO4.

Additional Examples

You need to make a 5%^w/v solution of CaCl₂, but only CaCl₂ · 2H₂O is available. How do you make the 5%^w/v solution?

To make this solution, use the formula for interchanging between anhydrous and hydrated chemicals:

You need to make a 25%^w/v solution of CaSO₄ but only CaSO4⋅2H2O is available. How do you make this solution?

To make this solution, use the formula for interchanging between anhydrous and hydrated chemicals:

You need to make a 15%^w/v solution of Na₂CO₃, but only Na2CO3⋅H2O is available. How do you make this solution?

To make this solution, use the formula for interchanging between anhydrous and hydrated chemicals:

A buffer solution needs to be prepared. The directions call for 2.50 g of anhydrous Na2HPO4 to be dissolved into 100 mL of deionized water. Only Na2HPO4⋅7H2O is available in the chemical storeroom. How much of the Na2HPO4⋅7H2O should be used to prepare the solution?

The molecular weight of Na2HPO4 is 141.96, whereas the molecular weight of the hydrated form is 268.10 because there are 7.00 water molecules attached to the salt. By substituting into the ratio formula the data from the problem, the following equation is derived:

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Tags: Mathematics for the Clinical Laboratory

Nov 18, 2017 | Posted by admin in PHARMACY | Comments Off

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