Análisis Estadístico

Specific gravity is the ratio of the density of a solution to the density of water at 4°C (1 g/mL). A substance with a value greater than 1 is denser than water, and a substance with a value smaller than 1 is less dense. And, because specific gravity is a ratio of two densities, the units cancel each other, leaving only a magnitude.

Specific gravity is helpful in diluting concentrated commercial acids in the laboratory. Measuring the weight of a liquid is awkward and, when that liquid is a concentrated acid, dangerous. Measuring its volume, however, is markedly easier, although precautions are still necessary.

Consider, for example, a 1 N (pronounced “one-normal”) HCl solution. By definition, this solution comprises 1 equivalent weight of HCl, or 36.5 g, in a volume of 1 L. The amount of HCl that contains 1 mole of hydrogen ions is, of course, 1 mole; the molar mass of HCl is 36.5 g, which is, therefore, the equivalent weight. Thus, for HCl the normality and molarity are equal.

By contrast, consider a 1 N H2SO4 solution. Although this solution, like 1 N HCl, comprises 1 equiv-

alent weight of H2SO4 in 1 L, only one-half a mole of H2SO4 contains 1 mole of hydrogen ions. Because

the molar mass of H2SO4 is 98.1 g, the equivalent weight is half that value, or 49.0 g. Thus, for H2SO4, a

1 N solution is the same as a 0.5 m solution.

For an acid, then, an equivalent is one hydrogen ion in the formula. Thus, 1 mole of HCl is 1 equiva- lent, whereas 1 mole of H2SO4 is 2 equivalents and 1 mole of H3PO4 is 3 equivalents.

For a base or a salt, an equivalent is the number of hydrogen ions with which it can theoretically combine; however, we can just as easily regard it as 1 mole of ionic charges. For example, because the bicarbonate ion (HCO3 -) carries a single charge and can combine with 1 hydrogen ion, then obviously

1 mole of bicarbonate ions carries 1 mole of charges and can combine with 1 mole of hydrogen ions. Therefore, 1 equivalent of HCO3 - ions is 1 mole.

The magnesium ion (Mg2+_{), by contrast, carries two charges and can theoretically replace two}

hydrogen ions. Thus, 1 mole of Mg2+ _{ions is the same as 2 equivalents.}

In general, then, the number of equivalents is the product of the number of moles and the number of charges:

equivalents = moles * charges

In fact, we can express concentration in terms of equivalents (“Eq”), a system widely used in medicine because it directly reports the concentration of positive or negative charges. For example, the concentra- tion of potassium ion in serum typically can be about 5 mEq/L, which is the same as 5 mmol/L (or 5 mm). However, calcium ion (Ca2+_{) can also typically be present in serum at about 5 mEq/L, which is not}

5 mmol/L, but half of that, or 2.5 mmol/L. Clearly, a mole of potassium ions gives the same concentra- tion of positive charges as does half a mole of calcium ions, but expressing that concentration in terms of equivalents makes it necessary to convert from moles to charges.

CheCkpoint 5-3

1. What is the normality of a solution consisting of 2 mol naCl in 1 liter? 2. A 2 N H2So4 solution has how many grams of H2So4?

1. 2 N. one mole of naCl is 1 equivalent.

2. An equivalent weight for sulfuric acid is half a mole, 49 g. Therefore, 2 equivalent weights is 98 g, the molar mass.

Suppose, for example, that you have a bottle of concentrated aqueous HCl, the label of which states the specific gravity to be 1.18 and the purity 36%. What these two values mean is that 1 mL of the liquid in the bottle weighs 1.18 g and that 36% of this weight is HCl (the rest is water). If your task is to dilute this acid to a concentration of 1 N in a final volume of 100 mL, then the final solution must have 1 equivalent weight of HCl in every liter.

To do this, you must first determine the equivalent weight of HCl in 100 mL of the dilute acid. Because HCl contains one hydrogen ion, the equivalent weight is the formula mass, or 36.5 g. Therefore,

100 mL ¢_{1000 mL}1 L ≤ ¢36.5 g_{1 L} ≤ = 3.65 g

What this means is that the diluted acid has 3.65 g of HCl in a final volume of 100 mL. Next, you must calculate the volume of concentrated aqueous HCl that contains 3.65 g—the volume that you will subsequently dilute to that final volume of 100 mL. You know that 36% of the weight of the concentrated aqueous acid is HCl; therefore,

0.36 * ¢1.18 g_{1 mL} ≤ = 0.425 g/mL

This means that every mL of the liquid contains 0.425 g of HCl. Now you can calculate the volume needed for dilution to 100 mL:

3.65 g HCl * ¢_{0.425 g HCl}1 mL ≤ = 8.6 mL

Thus, you pipet 8.6 mL (slowly and under a hood) of the concentrated aqueous HCl into water and then add enough water to bring the final volume to 100 mL. The result is a solution of 1 N HCl.

thE ph sCalE

The concentration of hydrogen ions profoundly affects many chemical reactions in the laboratory and nearly all physiological processes in the human body. Understanding the expression of hydrogen-ion concentrations, therefore, is critical both to the physician, who makes medical decisions, and to the laboratorian, who generates the test results that help guide those decisions.

Serum, urine, other biological fluids, and common laboratory solutions have H+ _{concentrations}

that are ponderous to express in the units discussed previously. For example, the concentration of H+ _{in the blood is 0.00000004 m (4 * 10}-8 _{m). To simplify such inconvenient numbers, Danish bio-}

chemist Søren Sørenson proposed the quantity “pH” in 1909. He defined the term pH as the puissance d’hydrogène, which translates as the power of hydrogen, expressing [H+_{] as a negative logarithm of 10. In}

other words, [H+_{] = 10}-pH_.

Rearranging this equation gives

pH = -log[H+_] obsolete Quantity

and symbol proposed term proposed symbol proposed Units

Molarity (m) Amount-of-substance concentration of B cB mol/dm3

mol/l kmol/m3

Molality (m) Molality of solute B bB mol/kg

normality (N) Amount-of-substance concentration of HnA

c [(1/n)HnA]* mol/dm3

mol/l kmol/m3

*In this notation, n represents the number of hydrogen ions an acid can release. For example, if the acid is sulfuric (H2So4), then rather than writing “a 0.5 N solution of sulfuric acid,” we would write “a solution of

sulfuric acid with an amount-of-substance concentration of c [(1/2) H2So4] of 0.5 mol/l.”

H tablE 5-1 Concentration Terms That NIST Considers Obsolete,

and Their Proposed Replacements

The pH of blood, then, is

pH = -log(4 * 10-8_m) pH = 7.4

Clearly, “7.4” is much simpler than “0.00000004” or “4 * 10-8_{.” Now, let us look at the pH scale}

(Table 5-2 H_{) that emerges from Equation 4.}

This scale has at least four salient features:

1. As pH increases, the concentration of H+ _{decreases. Thus, their relationship is inverse.}

2. A difference of 1 pH unit—for example, 8 to 9 or 5 to 4—represents a 10-fold change in the H+ _con-

centration. This is so because the scale is logarithmic and its base is 10. Thus, with respect to H+_{, a}

solution at pH 6 is 100 times more concentrated than it is at pH 8.

3. The concentration of H+ _{is in “mol/L.” The pH convention is valid only for this unit of concentration.}

4. By virtue of being a logarithm, the pH is dimensionless; that is, it has no units. Thus, we say “the pH is 7,” not “the pH is 7 moles per liter.”

A solution at pH 7 is “neutral.” At pH 6 7, it is acidic, and at pH 7 7, it is alkaline, or basic. (Strictly speaking, this is true only at 25°C. At other temperatures, these guidelines vary somewhat.)

Although a whole pH unit corresponds to a factor of 10, every decrement of 0.30 pH units reflects a doubling of the H+ _{concentration, and every increment reflects a halving of the H}+ _{concentration.}

This knowledge is helpful in a quick comparison of two pH values. If the H+ _{concentration doubles, the}

pH goes down by 0.30, and if the H+ _{concentration halves, the pH goes up by 0.30. For example, if the}

pH decreases from 4.50 to 4.20, the H+ _{concentration has risen by a factor of ∼2 from 3.16 * 10}-5 _m

to 6.31 * 10-5 _{m. Likewise, if the pH increases from 7.10 to 7.40, the H}+ _{concentration has fallen by a}

factor of ∼2 from 7.94 * 10-8 _{m to 3.98 * 10}-8 _m.

This relationship comes about because the logarithm of 2 is 0.30. Consider what happens when the H+ _{concentration doubles:}

pH = -log([H+_{]* 2)} pH = -log[H+_{] - log 2}

= -log[H+_{] - 0.30}

What the last equation in this sequence says is that when [H+_{] doubles, the original pH goes down by 0.30.}

When the H+ _{concentration halves, the original pH goes up by 0.30:}

[H+_{] (mol/l) pH} 0.1 10-1 ₁ acidic neutral alkaline 0.01 10-2 ₂ 0.001 10-3 ₃ 0.0001 10-4 ₄ 0.00001 10-5 ₅ 0.000001 10-6 6 0.0000001 10-7 ₇ 0.00000001 10-8 8 0.000000001 10-9 ₉ 0.0000000001 10-10 ₁₀ 0.00000000001 10-11 ₁₁ 0.000000000001 10-12 ₁₂ 0.0000000000001 10-13 ₁₃ 0.00000000000001 10-14 ₁₄