Acid-Base Indicators

What is an acid-base indicator? An acid-base indicator is a weak acid or a weak base. The undissociated form of the indicator is a different color than the iogenic form of the indicator. An Indicator does not change color from pure acid to pure alkaline at specific hydrogen ion concentration, but rather, color change occurs over a range of hydrogen ion concentrations. This range is termed the color change interval. It is expressed as a pH range.

How is an indicator used? Weak acids are titrated in the presence of indicators which change under slightly alkaline conditions. Weak bases should be titrated in the presence of indicators which change under slightly acidic conditions.

What are some common acid-base indicators? Several acid-base indicators are listed below, some more than once if they can be used over multiple pH ranges. Quantity of indicator in aqueous (aq.) or alcohol (alc.) solution is specified. Tried-and-true indicators include: thymol blue, tropeolin OO, methyl yellow, methyl orange, bromphenol blue, bromcresol green, methyl red, bromthymol blue, phenol red, neutral red, phenolphthalein, thymolphthalein, alizarin yellow, tropeolin O, nitramine, and trinitrobenzoic acid. Data in this table are for sodium salts of thymol blue, bromphenol blue, tetrabromphenol blue, bromcresol green, methyl red, bromthymol blue, phenol red, and cresol red.

IndicatorpH Range

Quantity per 10 ml Acid Base

Thymol Blue 1.2-2.8 1-2 drops 0.1% soln. in aq. red yellow

Pentamethoxy red 1.2-2.31 drop 0.1% soln. in 70% alc.

red-violet colorless

Tropeolin OO 1.3-3.2 1 drop 1% aq. soln. red yellow

2,4-Dinitrophenol 2.4-4.01-2 drops 0.1% soln. in 50% alc.

colorless yellow

Methyl yellow 2.9-4.01 drop 0.1% soln. in 90% alc.

red yellow

Methyl orange 3.1-4.4 1 drop 0.1% aq. soln. red orange

Bromphenol blue 3.0-4.6 1 drop 0.1% aq. soln. yellow blue-violet

Tetrabromphenol blue 3.0-4.6 1 drop 0.1% aq. soln. yellow blue

Alizarin sodium sulfonate

3.7-5.2 1 drop 0.1% aq. soln. yellow violet

α-Naphthyl red 3.7-5.01 drop 0.1% soln. in 70% alc.

red yellow

p-Ethoxychrysoidine 3.5-5.5 1 drop 0.1% aq. soln. red yellow

Bromcresol green 4.0-5.6 1 drop 0.1% aq. soln. yellow blue

Methyl red 4.4-6.2 1 drop 0.1% aq. soln. red yellow

Bromcresol purple 5.2-6.8 1 drop 0.1% aq. soln. yellow purple

Chlorphenol red 5.4-6.8 1 drop 0.1% aq. soln. yellow red

Bromphenol blue 6.2-7.6 1 drop 0.1% aq. soln. yellow blue

p-Nitrophenol 5.0-7.0 1-5 drops 0.1% aq. soln. colorless yellow

Azolitmin 5.0-8.0 5 drops 0.5% aq. soln. red blue

Phenol red 6.4-8.0 1 drop 0.1% aq. soln. yellow red

Neutral red 6.8-8.01 drop 0.1% soln. in 70% alc.

red yellow

Rosolic acid 6.8-8.01 drop 0.1% soln. in 90% alc.

yellow red

Cresol red 7.2-8.8 1 drop 0.1% aq. soln. yellow red

α-Naphtholphthalein 7.3-8.71-5 drops 0.1% soln. in 70% alc.

rose green

Tropeolin OOO 7.6-8.9 1 drop 0.1% aq. soln. yellow rose-red

Thymol blue 8.0-9.6 1-5 drops 0.1% aq. soln. yellow blue

Phenolphthalein 8.0-10.01-5 drops 0.1% soln. in 70% alc.

colorless red

α-Naphtholbenzein 9.0-11.01-5 drops 0.1% soln. in 90% alc.

yellow blue

Thymolphthalein 9.4-10.61 drop 0.1% soln. in 90% alc.

colorless blue

Nile blue 10.1-11.1 1 drop 0.1% aq. soln. blue red

Alizarin yellow 10.0-12.0 1 drop 0.1% aq. soln. yellow lilac

Salicyl yellow 10.0-12.01-5 drops 0.1% soln. in 90% alc.

yelloworange-brown

Diazo violet 10.1-12.0 1 drop 0.1% aq. soln. yellow violet

Tropeolin O 11.0-13.0 1 drop 0.1% aq. soln. yelloworange-brown

Nitramine 11.0-13.01-2 drops 0.1% soln in 70% alc.

colorlessorange-brown

Poirrier's blue 11.0-13.0 1 drop 0.1% aq. soln. blue violet-pink

Trinitrobenzoic acid 12.0-13.4 1 drop 0.1% aq. soln. colorless orange-red

The concentration of a chemical solution refers to the amount of solute that is dissolved in a solvent. We normally think of a solute as a solid that is added to a solvent (e.g., adding table salt to water), but the solute could just as easily exist in another phase. For example, if we add a small amount of ethanol to water, then the ethanol is the solute and the water is the solvent. If we add a smaller amount of water to a larger amount of ethanol, then the water could be the solute!

Units of Concentration

Once you have identified the solute and solvent in a solution, you are ready to determine its concentration. Concentration may be expressed several different ways, using percent composition by mass, volume percent, mole fraction, molarity, molality, or normality.

1. Percent Composition by Mass (%)

This is the mass of the solute divided by the mass of the solution (mass of solute plus mass of solvent), multiplied by 100.

Example:Determine the percent composition by mass of a 100 g salt solution which contains 20 g salt.

Solution:20 g NaCl / 100 g solution x 100 = 20% NaCl solution

2. Volume Percent (% v/v) Volume percent or volume/volume percent most often is used when preparing solutions of liquids. Volume percent is defined as:

v/v % = [(volume of solute)/(volume of solution)] x 100%

Note that volume percent is relative to volume of solution, not volume of solvent. For example, wine is about 12% v/v ethanol. This means there are 12 ml ethanol for every 100 ml of wine. It is important to realize liqud and gas volumes are not necessarily additive. If you mix 12 ml of ethanol and 100 ml of wine, you will get less than 112 ml of solution.

As another example. 70% v/v rubbing alcohol may be prepared by taking 700 ml of isopropyl alcohol and adding sufficient water to obtain 1000 ml of solution (which will not be 300 ml).

3. Mole Fraction (X) This is the number of moles of a compound divided by the total number of moles of all chemical species in the solution. Keep in mind, the sum of all mole fractions in a solution always equals 1.

Example:What are the mole fractions of the components of the solution formed when 92 g glycerol is mixed with 90 g water? (molecular weight water = 18; molecular weight of glycerol = 92)

Solution:90 g water = 90 g x 1 mol / 18 g = 5 mol water92 g glycerol = 92 g x 1 mol / 92 g = 1 mol glyceroltotal mol = 5 + 1 = 6 molxwater = 5 mol / 6 mol = 0.833x glycerol = 1 mol / 6 mol = 0.167It's a good idea to check your math by making sure the mole fractions add up to 1:xwater + xglycerol = .833 + 0.167 = 1.000

4. Molarity (M)

Molarity is probably the most commonly used unit of concentration. It is the number of moles of solute per liter of solution (not necessarily the same as the volume of solvent!).

Example:What is the molarity of a solution made when water is added to 11 g CaCl2 to make 100 mL of solution?

Solution:11 g CaCl2 / (110 g CaCl2 / mol CaCl2) = 0.10 mol CaCl2

100 mL x 1 L / 1000 mL = 0.10 Lmolarity = 0.10 mol / 0.10 Lmolarity = 1.0 M

5. Molality (m)

Molality is the number of moles of solute per kilogram of solvent. Because the density of water at 25°C is about 1 kilogram per liter, molality is approximately equal to molarity for dilute aqueous solutions at this temperature. This is a useful approximation, but remember that it is only an approximation and doesn't apply when the solution is at a different temperature, isn't dilute, or uses a solvent other than water.

Example:What is the molality of a solution of 10 g NaOH in 500 g water?

Solution:10 g NaOH / (40 g NaOH / 1 mol NaOH) = 0.25 mol NaOH500 g water x 1 kg / 1000 g = 0.50 kg watermolality = 0.25 mol / 0.50 kgmolality = 0.05 M / kgmolality = 0.50 m

6. Normality (N)

Normality is equal to the gram equivalent weight of a solute per liter of solution. A gram equivalent weight or equivalent is a measure of the reactive capcity of a given molecule. Normality is the only concentration unit that is reaction dependent.

Example:1 M sulfuric acid (H2SO4) is 2 N for acid-base reactions because each mole of sulfuric acid provides 2 moles of H+ ions. On the other hand, 1 M sulfuric acid is 1 N for sulfate precipitation, since 1 mole of sulfuric acid provides 1 mole of sulfate ions.

Dilutions

You dilute a solution whenever you add solvent to a solution. Adding solvent results in a solution of lower concentration. You can calculate the concentration of a solution following a dilution by applying this equation:

MiVi = MfVf

where M is molarity, V is volume, and the subscripts i and f refer to the initial and final values.

Example:How many millilieters of 5.5 M NaOH are needed to prepare 300 mL of 1.2 M NaOH?

Solution:5.5 M x V1 = 1.2 M x 0.3 LV1 = 1.2 M x 0.3 L / 5.5 MV1 = 0.065 LV1 = 65 mL

So, to prepare the 1.2 M NaOH solution, you pour 65 mL of 5.5 M NaOH into your container and add water to get 300 mL final volume.

ALLOYAn alloy is a substance made by melting two or more elements together, at least one of them a metal. An alloy crystallizes upon cooling into a solid solution, mixture, or intermetallic compound.Examples of alloys include brass, bronze, 14k gold, and sterling silver

What Is Distillation

Distillation is a widely used method for separating mixtures based on differences in the conditions required to change the phase of components of the mixture. To separate a mixture of liquids, the liquid can be heated to force components, which have different boiling points, into the gas phase. The gas is then condensed back into liquid form and collected. Repeating the process on the collected liquid to improve the purity of the product is called double distillation. Although the term is most commonly applied to liquids, the reverse process can be used to separate gases by liquefying components using changes in temperature and/or pressure.

Distillation is used for many commercial processes, such as production of gasoline, distilled water, xylene, alcohol, paraffin, kerosene, and many other liquids. Types of distillation include simple distillation (described here), fractional distillation (different volatile 'fractions' are collected as they are produced), and destructive distillation (usually, a material is heated so that it decomposes into compounds for collection).

Chemistry of Hard and Soft Water

I'm sure you've heard the terms 'hard water' and 'soft water', but do you know what they mean? Is one type of water somehow better than the other? What type of water do you have? Let's take a look at the definitions of these terms and how they relate to water in everyday life.

Hard water is any water containing an appreciable quantity of dissolved minerals. Soft water is treated water in which the only cation (positively charged ion) is sodium. The minerals in water give it a characteristic taste. Some natural mineral waters are highly sought for their flavor and the health benefits they may confer. Soft water, on the other hand, may taste salty and may not be suitable for drinking.

If soft water tastes bad, then why might you use a water softener? The answer is that extremely hard water may shorten the life of plumbing and lessen the effectiveness of certain cleaning agents. When hard water is heated, the carbonates precipitate out of solution, forming scale in pipes and tea kettles. In addition to narrowing and potentially clogging the pipes, scale prevents efficient heat transfer, so a water heater with scale will have to use a lot of energy to give you hot water. Soap is less effective in hard water because its reacts to form the calcium or magnesium salt of the organic acid of the soap. These salts are insoluble and form grayish soap scum, but no cleansing lather. Detergents, on the other hand, lather in both hard and soft water. Calcium and magnesium salts of the detergent's organic acids form, but these salts are soluble in water.

Hard water can be softened (have its minerals removed) by treating it with lime or by passing it over an ion exchange resin. The ion exchange resins are complex sodium salts. Water flows over the resin surface, dissolving the sodium. The calcium, magnesium, and other cations precipitate onto the resin surface. Sodium goes into the water, but the other cations stay with the resin. Very hard water will end up tasting saltier than water that had fewer dissolved minerals.

Most of the ions have been removed in soft water, but sodium and various anions (negatively charged ions) still remain. Water can be deionized by using a resin that replaces cations with hydrogen and anions with hydroxide. With this type of resin, the cations stick to the resin and the hydrogen and hydroxide that are released combine to form pure water.

Salt Formation

When acids and bases react with each other, they can form a salt and (usually) water. This is called a neutralization reaction and takes the following form:

HA + BOH --> BA + H2O

Depending on the solubility of the salt, it may remain in ionized form in the solution or it may precipitate out of solution. Neutralization reactions usually proceed to completion.

The reverse of the neutralization reaction is called hydrolysis. In a hydrolysis reaction a salt reacts with water to yield the acid or base:

BA + H2O --> HA + BOH

More specifically, there are four combinations of strong and weak acids and bases:

strong acid + strong base, e.g., HCl + NaOH --> NaCl + H2O

When strong acids and strong bases react, the products are salt and water. The acid and base neutralize each other, so the solution will be neutral (pH=7) and the ions that are formed will not reaction with the water.

strong acid + weak base, e.g., HCl + NH3 ---> NH4Cl

The reaction between a strong acid and a weak base also produces a salt, but water is not usually formed because weak bases tend not to be hydroxides. In this case, the water solvent will react with the cation of the salt to reform the weak base. For example:

HCl (aq) + NH3 (aq) <--> NH4+ (aq) + Cl- while

NH4- (aq) + H2O <--> NH3 (aq) + H3O+ (aq)

weak acid + strong base, e.g., HClO + NaOH --> NaClO + H2O

When a weak acid reacts with a strong base the resulting solution will be basic. The salt will be hydrolyzed to form the acid, together with the formation of the hydroxide ion from the hydrolyzed water molecules.

weak acid + weak base, e.g., HClO + NH3 <--> NH4ClO

The pH of the solution formed from the reaction of a weak acid with a weak base depends on the relative strengths of the reactants. For example, if the acid HClO has a Ka of 3.4 x 10-8 and the base NH3 has a Kb = 1.6 x 10-5, then the aqueous solution of HClO and NH3 will be basic because the Ka of HClO is less than the Ka of NH3.