Chemistry
pH
pH is a measure of the acidity or basicity of a solution. It is defined as the negative logarithm of the concentration of hydrogen ions in the solution. A pH of 7 is considered neutral, while values below 7 are acidic and values above 7 are basic.
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eBook - PDFAnalytical Chemistry
A Toolkit for Scientists and Laboratory Technicians
- Bryan M. Ham, Aihui MaHam(Authors)
- 2024(Publication Date)
- Wiley(Publisher)
The pH scale in general is defined as: If the solution pH is 7.0, the solution is neutral. If the solution pH is <7.0, the solution is acidic. If the solution pH is >7.0, the solution is basic. The pOH in Figure 9.4 is the negative log of the molar [OH − ] concentration. pOH = − log[OH − ] (9.61) A useful relationship exists between pH, pOH, and the ion product for water, K w : pH + pOH = pK w (9.62) MEASURING THE pH 171 Acidic pH Basic pH pH Scale 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 pH [H + ] [OH – ] pOH 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 1 × 10 –14 1 × 10 0 1 × 10 –13 1 × 10 –12 1 × 10 –11 1 × 10 –10 1 × 10 –9 1 × 10 –8 1 × 10 –7 1 × 10 –6 1 × 10 –5 1 × 10 –4 1 × 10 –3 1 × 10 –2 1 × 10 –1 1 × 10 0 1 × 10 –14 1 × 10 –1 1 × 10 –2 1 × 10 –3 1 × 10 –4 1 × 10 –5 1 × 10 –6 1 × 10 –7 1 × 10 –8 1 × 10 –9 1 × 10 –10 1 × 10 –11 1 × 10 –12 1 × 10 –13 Ammonia solution Baking soda Milk Blood Rain water Coffee Stomach fluid Soft drink Fresh egg Lemons Tomatoes Calcium carbonate Milk of magnesia FIGURE 9.4 The pH scale, the hydrogen ion concentration [H + ], the hydroxide ion concentration [OH − ], the pOH, and the pH of some common solutions and substances. If we know the pH of a solution, then we can calculate the pOH from the relationship in Equation 9.12. Suppose a solution has a pH of 5.9. The value of pK w is −log(1 × 10 −14 ) = 14, thus: 5.9 + pOH = 14 (9.63) pOH = 8.1 (9.64) 9.13 MEASURING THE pH The laboratory technician and chemical analyst will often mea- sure the pH of a solution or of a substance dissolved in water. The pH is not usually calculated from a determination of the hy- drogen ion [H + ] molar concentration by any wet chemical anal- yses such as titration, but typically by a pH meter, or sometimes as an approximate estimation by pH paper or strips as depicted in Figure 9.2. An example of a typical pH meter is depicted in Figure 9.5. The pH meter system consists of a pH-sensing elec- trode and a readout screen.
eBook - PDF- John Kenkel(Author)
- 2010(Publication Date)
- CRC Press(Publisher)
In pure, dis-tilled water, or in a neutralized water solution, the [H + ] and the [OH − ] are equal to 1 .0 × 10 − 7 M . The pH of a neutral solution, such as pure, distilled water, is thus 7 .00 . Any pH values between 0 and 7 indicate an acidic solution, one in which hydrogen ions predominate over hydroxide ions . Any pH values between 7 and 14 indicate a basic solution, one in which hydroxide ions pre-dominate over hydrogen ions . We see from this that the relationship between [H + ] and pH is an inverse one; as [H + ] increases, pH decreases, and vice versa . A low pH indicates a high [H + ] (compared with [OH − ]), while a high pH indi-cates a low [H + ] . How is the logarithm of a number determined? Since every positive num-ber has a logarithm specific to that number, tables of logarithms have been created, and logarithms can be found by looking at these tables . However, logarithms can also be determined by using the LOG key on a scientific cal-culator . To determine the pH of a solution given a hydrogen ion concentration that is expressed in scientific notation, you will need to learn the required keystrokes on your calculator . Here is an example for practice: What is the pH of a solution that has an [H + ] of 4 .59 × 10 − 5 M? The answer is 4 .338 . Table 12.5 Examples of Logarithms of Numbers That Are Not Multiples of 10 10 0.30103 = 2 log 2 = 0.30103 10 0.4771213 = 3 log 3 = 0.4771213 10 0.7092123 = 5.11932 log 5.11932 = 0.7092123 Basic Chemistry Concepts and Exercises 314 It is also important to be able to calculate the [H + ] given the pH . Here is an example of this for practice: If the pH of a solution is 3 .51, what is the [H + ]? The answer is 3 .1 × 10 − 4 M .
eBook - PDF- Gary D. Christian, Purnendu K. Dasgupta, Kevin A. Schug(Authors)
- 2013(Publication Date)
- Wiley(Publisher)
ranges, from 1 M or greater to 10 −14 M or less. To construct a plot of H + concentration against some variable would be very difficult if the concentration changed from, say, 10 −1 M to 10 −13 M. This range is common in a titration. It is more convenient to 228 CHAPTER 7 ACID–BASE EQUILIBRIA compress the acidity scale by placing it on a logarithm basis. The pH of a solution was defined by Sørenson as pH = −log[H + ] (7.15) The minus sign is used because most of the concentrations encountered are less than pH is really –log a H + . This is what a pH meter (glass electrode) measures—see Chapter 13. 1 M, and so this designation gives a positive number. (More strictly, pH is now defined as − log a H + , but we will use the simpler definition of Equation 7.15.) In general, pAnything = − log Anything, and this method of notation will be used later for other numbers that can vary by large amounts, or are very large or small (e.g., equilibrium Carlsberg Laboratory archives In 1909, Søren Sørenson, head of the chemistry department at Carlsberg Laboratory (Carlsberg Brewery) invented the term pH to describe this effect and defined it as −log[H + ]. The term pH refers simply to “the power of hydrogen.” In 1924, he realized that the pH of a solution is a function of the “activity” of the H + ion, and published a second paper on the subject, defining it as pH = − log a H + . constants). Example 7.2 Calculate the pH of a 2.0 × 10 −3 M solution of HCl. Solution HCl is completely ionized, so [H + ] = 2.0 × 10 −3 M pH = − log(2.0 × 10 −3 ) = 3 − log 2.0 = 3 − 0.30 = 2.70 A similar definition is made for the hydroxide ion concentration: pOH = −log[OH − ] (7.16) Equation 7.13 can be used to calculate the hydroxyl ion concentration if the A 1 M HCl solution has a pH of 0 and pOH of 14. A 1 M NaOH solution has a pH of 14 and a pOH of 0. hydrogen ion concentration is known, and vice versa.
eBook - ePubSoil Science
Methods & Applications
- David L. Rowell(Author)
- 2014(Publication Date)
- Routledge(Publisher)
Section 8.1 explains the terms involved. It should be noted that when applied to soils, ‘neutral’ is given a slightly different meaning, being a range from about pH 6.5 to 7.Soil acidity involves more than just the pH of the soil solution. This is still the main principle and the measurement of soil pH (Section 8.1 ) is normally made in a suspension of soil in water such that the value obtained is primarily related to the solution pH. However, hydrogen ions are also present on cation exchange sites and have an effect on the measurement. Also as soils become more acidic (pH 7 → 3), there are associated changes in the following properties:• The amounts of exchangeable Ca2+ and Mg2+ decrease. These together with exchangeable K+ , Na2+ and are known as the basic cations: their total amount is often expressed as a percentage of the CEC which is termed the percentage base saturation (Section 8.2 ).• The amount of exchangeable Al3+ increases and is often expressed as the percentage aluminium saturation of the ECEC (Section 8.2 ).• The negative charge on humus decreases and the positive charge on sesquioxides increases (Sections 7.1 and 7.5 ).• The availability of plant nutrients is changed. For example, pHospHate solubility is reduced (Ch. 10 ).• The availability of toxic elements is changed. For example, aluminium and manganese become more soluble in acid soils (Section 8.3 ).• The activity of many soil organisms is reduced resulting in an accumulation of organic matter, reduced mineralization and a lower availability of N, P and S. THE DEVELOPMENT OF SOIL ACIDITYIn pure water the concentration of H+ ions is 10−7 mol 1−1 and the pH is 7. When in contact with the atmospHeric concentration of CO2 a dilute carbonic acid solution is formed with a pH of 5.6. Distilled or deionized water in the laboratory therefore has a pH of about 5.6. For the pH to differ from this value some other acid or base must be present. Thus ‘acid rain’ contains nitric and sulpHuric acid dissolved from the atmospHere (or ammonia and oxides of N and S which can form these acids). Its pH is below 5.6; the average pH of rain over eastern Britain is about 4.4 (DOE, 1990). Even in unpolluted air rain picks up small amounts of naturally occurring acid and has a pH of about 5. Ammonia and oxides of N and S are also deposited dry on vegetation and soil and are washed into the soil by rain where they produce acidity. Thus the atmospHere is an external source of acidity (Fig. 8.1
eBook - PDF- David Ucko(Author)
- 2012(Publication Date)
- Academic Press(Publisher)
T h e p H of a solution is a direct measure o f its hydrogen ion concentration. It is a scale running from 0, strongly acidic, to 14, strongly basic, in water. The midpoint, p H 7, represents a neutral solution, neither acidic nor basic. An acid has a p H b e l o w 7, while a base has a p H above 7. T h e p H of a solution can b e determined using certain dyes, called indicators, or b y a p H meter. Neutralization is the reaction of an acid with a base. O n e hydroxide ion ex-actly cancels out the effect of one hydrogen ion, forming a neutral water molecule. Antacids work b y neutralizing acids in this way. A titration is a neu-tralization reaction performed to find the concentration o f an acid or a base. An equivalent or equivalent weight of an acid is the weight in grams that provides 1 mole o f hydrogen ions. For a base, it is the weight that reacts with 1 mole of hydrogen ions. Normality is defined as the number of equivalents of acid or base per liter o f solution. T h e normality of a solution is always greater than or equal to its molarity. A salt, an ionic c o m p o u n d , forms w h e n an acid and base react. It consists of a cation from the base and an anion from the acid. Some salts react with water in a process called hydrolysis, changing the p H from the neutral value of 7. T h e salts present in the b o d y exist in solution as ions and are therefore called electrolytes. T h e most important electrolytes are N a + , K + , C a 2 + , Mg 2 1 , CI, H C 0 3 ~ , H P 0 4 2 ~ , and S 0 4 2 . T h e y are essential for maintaining fluid bal-ance and acid-base balance, as well as for the normal functioning of the cells. Their concentrations are expressed as milliequivalents per liter. 174 a ter ds ases and sa ts Buffers keep the p H values of the b o d y fluids within their normal range. T h e y keep the p H from changing b y neutralizing small amounts of acid or base that may b e added to the system.
eBook - PDFChemistry
The Molecular Nature of Matter
- James E. Brady, Neil D. Jespersen, Alison Hyslop(Authors)
- 2014(Publication Date)
- Wiley(Publisher)
According to the color code, the pH of the solution is closer to 3 than to the color for pH 5. Andy Washnik 806 Chapter 16 | Acid–Base Equilibria in Aqueous Solutions | Summary Organized by Learning Objective Define pH and explain the use of “p” notation Water reacts with itself to produce small amounts of H 3 O + (often abbreviated H + ) and OH - ions. The concentrations of these ions, both in pure water and dilute aqueous solutions, are related by the expression 3 H + 4 3 OH - 4 = K w = 1.0 Ž 10 -14 (at 25 °C) K w is the ion product constant of water. In pure water 3 H + 4 = 3 OH - 4 = 1.0 Ž 10 -7 The pH of a solution is defined by the equation, pH = -log[H + ]. As the pH decreases, the acidity, or [H + ], increases. The compa- rable term, pOH (= -log[OH - ]), is used to describe a solution that is basic. A solution is acidic if the hydrogen ion concentration exceeds 1.0 Ž 10 -7 or the pH is less than 7.00. Similarly, a solution is basic if the hydroxide ion concentration exceeds 1.0 Ž 10 -7 or if the pH is greater than 7.00. Explain how to determine the pH of strong acids or bases in aqueous solution When calculating the pH of strong acids or strong bases, we assume that they are 100% ionized. Write expressions for the acid ionization constant, K a , and base ionization constant, K b , and explain how they are related to each other A weak acid H A ionizes according to the general equation H A + H 2 O m H 3 O + + A - or more simply, H A mH + + A - The equilibrium constant is called the acid ionization con- stant, K a : K a = 3 H 3 O + 4 3 A - 4 3 H A 4 A weak base B ionizes by the general equation B + H 2 O mBH + + OH - The equilibrium constant is called the base ionization con- stant, K b : K b = 3 BH + 4 3 OH - 4 3 B 4 The smaller the values of K a (or K b ), the weaker are the sub- stances as Brønsted acids (or bases).
eBook - PDF- Douglas P. Heller, Carl H. Snyder(Authors)
- 2015(Publication Date)
- Wiley(Publisher)
The pH Scale 243 pH: The Measure of Acidity As we just saw, we can make things simpler for ourselves by writing the value of [H 3 O + ] as an exponent of 10. For example, a hydronium ion concentration of 0.0000001 M is the same as 1 × 10 −7 M, which is simply 10 −7 M. We can carry this a step further by dispensing with both the 10 and the negative sign. This is exactly what the Danish biochemist Søren Sørensen did in 1909 when he proposed that concentrations of H + (or, as we now know, H 3 O + ) be treated as exponential values. (Mathematically, this process of using the exponent or power of a number as a value is called “taking the logarithm” of that num- ber.) Following Sørensen’s recommendation, we now consider the concentration of hydro- nium ion, [H 3 O + ], in terms of pH (Figure 8.10). The letters pH represent the “power of the Hydrogen (or Hydronium) ion.” As a symbol for acidity, pH reflects nicely the international character of chemistry. The let- ter p begins the English word power as well as its French and German equivalents, puissance and Potenz. At the time of Sørensen’s suggestion, English, French, and German were the world’s dominant scientific languages. In chemically pure water, the concentrations of all the transient hydronium (H 3 O + ) and hydroxide (OH − ) ions that exist at equilibrium are not only fixed at any given temperature but are also equal to each other. They must always equal each other in pure water be- cause the ionization of each water molecule produces an equal number of ions—one hydronium ion and one hydroxide ion. Experimental measurements show that each of these ions is present in pure, neutral water at a concentration of 0.0000001 moles per li- ter at 25°C. For brevity, we write [H 3 O + ] for “the molar concentration of H 3 O + ” and we express the value of the molar concentration in exponential notation, using the italicized capital M as the symbol for moles/liter.
- Peter V. Hobbs(Author)
- 2000(Publication Date)
- Cambridge University Press(Publisher)
We see from definition (5.14) that (1) the greater the hydrogen ion concentration (i.e., the more acidic the solution) the smaller is the pH value of the solution, and (2) a change in the hydrogen ion concentration by a factor of ten (e.g., from 10 1 to 10~ 2 M) changes the pH value by unity. At the beginning of this section we defined a solution as being neutral if [H + (aq)] = [OH(aq)]. Pure water is neutral; therefore, from Eqs. (5.12) and (5.13) [H 3 O + (aq)][OH-(aq)] = l(r 14 or, [H 3 O + (aq)] 2 =10-Therefore, for pure water [H 3 O + (aq)] = [H + (aq)] = 10 7 M Hence, the pH of pure water is -log(10~ 7 ) = 7. It follows that acidic solu-tions have pH < 7 and basic solutions have pH > 7. Observed pH values in nature are generally between 4 and 9. Seawater normally has a pH between 8.1 and 8.3. Streams in wet climates generally have a pH between 5 and 6.5 and in dry climates between 7 and 8. Soil water in the presence of abundant decaying vegetation may have a pH of 4 or lower. The pH of rainwater can range from quite acidic (around 4.0) in industrial regions to about 5.6 in very clean regions. We will discuss the acidity of rainwater in some detail at the end of this chapter, but the following exercise illustrates why even clean rainwater does not have a pH of 7. Exercise 5.2. The pH of natural rainwater is about 5.6. Assum-ing that all of this acidity is due to the absorption of CO 2 by the rain, determine how many moles of CO 2 would have to be absorbed in 1L of rainwater. Solution. Since the pH of rainwater is 5.6, the concentration of H 3 O + (aq) in natural rainwater is given by pH = 5.6 = -log[H 3 O + (aq)] Therefore, [H 3 O + (aq)] = 0.25xl0-5 M 90 Acids and bases The main source of H 3 O + (aq) when CO 2 dissolves in water is CO 2 (g) + H 2 O(l)+±H 2 CO 3 (aq) (5.15a) H 2 CO 3 (aq) + H 2 O(1) ?± HCO 3 -(aq) + H 3 O + (aq) (5.15b) We see from Reactions (5.15) that for every mole of CO 2 that is absorbed in water, one mole of H 3 O + (aq) is produced.
eBook - PDF- Gary D. Christian, Purnendu K. Dasgupta, Kevin A. Schug(Authors)
- 2020(Publication Date)
- Wiley(Publisher)
To construct a plot of H + concentra- tion against some variable would be very difficult if the concentration changed from, say, 10 -1 M to 10 -13 M. This range is common in a titration. It is more convenient to compress the acidity scale by placing it on a logarithm basis. The pH of a solution was defined by Sørenson as pH = - log[H + ] (6.15) Pdf_Folio:212 6.4 THE pH SCALE 213 The minus sign is used because most of the concentrations encountered are less than 1 M, and so this designation gives a positive number. (More strictly, pH is now defined as - log a H + , but we will use the simpler definition of Equation 6.15.) In general, pAnything = -log Anything, and this method of notation will be used later for other numbers that can vary by large amounts, or are very large or small (e.g., equilibrium constants). pH is really –log a H + . This is what a pH meter (glass electrode) measures—see Chapter 20. Example 6.2 Calculate the pH of a 2.0 × 10 -5 M solution of HCl. Solution HCl ion completely ionized, so [H + ] = 2.0 × 10 -5 M pH = - log(2.0 × 10 -5 ) = 5 - log 2.0 = 5 - 0.30 = 4.70 A similar definition is made for the hydroxide ion concentration: pOH = -log[OH - ] (6.16) Equation 6.13 can be used to calculate the hydroxyl ion concentration if the hydrogen ion concentration is known, and vice versa. The equation in logarithm form for a more direct calculation of pH or pOH is - log K w = - log[H + ][OH - ] = - log[H + ] - log[OH - ] (6.17) pK w = pH + pOH (6.18) A 1 M HCl solution has a pH of 0 and pOH of 14. A 1 M NaOH solution has a pH of 14 and a pOH of 0. At 25°C, 14.00 = pH + pOH (6.19) Example 6.3 Calculate the pOH and the pH of a 5.0 × 10 -5 M solution of NaOH at 25°C. Solution [OH - ] = 5.0 × 10 -5 M pOH = - log(5.0 × 10 -5 ) = 5 - log 5.0 = 5 - 0.70 = 4.30 pH + 4.30 = 14.00 pH = 9.70 or [H + ] = 1.0 × 10 -14 5.0 × 10 -5 = 2.0 × 10 -10 M pH = - log(2.0 × 10 -10 ) = 10 - log 2.0 = 10 - 0.30 = 9.70 Pdf _Folio:2 13
eBook - PDF- Charlotte W. Pratt, Kathleen Cornely(Authors)
- 2020(Publication Date)
- Wiley(Publisher)
2.3 Acid–Base Chemistry • The dissociation of water produces hydroxide ions (OH − ) and protons (H + ) whose concentration can be expressed as a pH value. The pH of a solution can be altered by adding an acid (which donates protons) or a base (which accepts protons). • The tendency for a proton to dissociate from an acid is expressed as a pK value. • The Henderson–Hasselbalch equation relates the pH of a solution of a weak acid and its conjugate base to the pK and the concentrations of the acid and base. 2.4 Tools and Techniques: Buffers • A buffered solution, which contains an acid and its conjugate base, resists changes in pH when more acid or base is added. 2.5 Clinical Connection: Acid–Base Balance in Humans • The body uses the bicarbonate buffer system to maintain a con- stant internal pH. Homeostatic adjustments are made by the lungs, where CO 2 is released, and by the kidneys, which excrete H + and ammonia. SUMMARY polarity hydrogen bond ionic interaction van der Waals radius electronegativity van der Waals interaction dipole–dipole interaction London dispersion forces dielectric constant solute solvation hydration KEY TERMS 46 CHAPTER 2 Aqueous Chemistry Brief Bioinformatics Exercises 2.1 Structure and Solubility 2.2 Amino Acids, Ionization, and pK Values BIOINFORMATICS 2.1 Water Molecules and Hydrogen Bonds 1. Each CO bond in CO 2 is polar, yet the whole molecule is nonpolar. Explain. 2. The HCH bond angle in the perfectly tetrahedral CH 4 molecule is 109º. Explain why the HO H bond angle in water is only about 104.5º. 3. Which compound has a higher boiling point, H 2 O or H 2 S? Explain. 4. Consider the following molecules and their melting points listed below. How can you account for the differences in melting points among these molecules of similar size? Molecular weight Melting (g · mol –1 ) point (°C) Water, H 2 O 18.0 0 Ammonia, NH 3 17.0 −77 Methane, CH 4 16.0 −182 5.
eBook - PDFGeneral, Organic, and Biological Chemistry
An Integrated Approach
- Kenneth W. Raymond(Author)
- 2012(Publication Date)
- Wiley(Publisher)
7.6 The pH Scale 253 SOLUTION a. [H 3 O + ] = 10 - pH = 10 - 6 = 1 * 10 - 6 M (acidic) b. [H 3 O + ] = 10 - pH = 10 - 6.5 = 3 * 10 - 7 M (acidic) c. [H 3 O + ] = 10 - pH = 10 - 1.2 = 6 * 10 - 2 M (acidic) Solving part a does not require a calculator. All three answers are reported with one signifi- cant figure (Math Support—Logs and Antilogs). PRACTICE PROBLEM 7.8 What is [OH - ] in the following solutions? Indicate whether each is acidic, neutral, or basic. a. pH = 7.2 b. pH = 9.1 c. pH = 3.3 MATH SUPPORT—LOGS AND ANTILOGS On your scientific calculator you should be able to calculate logarithms (logs). If you try a few calculations, you will see that log 100 = 2 log 1000 = 3 log 0.0001 = -4 Converting these three numbers (100, 1000, and 0.0001) into scientific notation (Section 1.4) should help explain what the log or logarithm of a number is. log 100 = log 10 2 = 2 log 1000 = log 10 3 = 3 log 0.0001 = log 10 -4 = -4 The log of a number is the power to which ten must be raised to equal the number (log 10 n = n). The log of 100 is 2, because 10 2 equals 100 and the log of 0.0001 is -4 because 10 -4 = 0.0001. When a number has the value 1 * 10 n , where n is an integer (-2, 5, etc.), its log can be determined without using a calculator. In these cases, the log is equal to the value of n. log 1 * 10 -2 = -2 log 1 * 10 5 = 5 In all other cases (7.9 * 10 2 , 2.2 * 10 -5 , etc.), a calculator will be required. log 7.9 * 10 2 = 2.90 log 2.2 * 10 -5 = -4.66 Reversing this process gives the antilog or antilogarithm of a number (antilog n = 10 n ). log of 10 2 = 2, so antilog 2 = 10 2 log of 10 -8 = -8, so antilog -8 = 10 -8 When n is an integer (2, -8, etc.), its antilog can be determined without using a calculator, because the antilog is equal to 10 n (see the two examples directly above). At other times (3.5, -1.3, etc.) a calculator must be used.
eBook - PDFPhysical Chemistry
Understanding our Chemical World
- Paul M. S. Monk(Author)
- 2005(Publication Date)
- Wiley(Publisher)
To make such a solution, we could calculate exactly how many moles of acid to add to water, but this method is generally difficult, since even small errors in weighing the acid can cause wide fluctuations in the pH. Furthermore, we cannot easily weigh out one of acid oxides such as NO. Anyway, the pH of a weak acid does not clearly follow the acid’s concentration (see p. 254). The Henderson–Hasselbach equation, Equation (6.50), relates In some texts, Equa- tion (6.50) is called the Henderson– HasselbaLch equation. the pH of a buffer solution to the amounts of conjugate acid and conjugate base it contains: pH = pK a + log 10 [A − ] [HA] (6.50) We follow the usual pattern here by making a buffer with a weak acid HA and a solution of its conjugate base, such as the sodium salt of the respective anion, A − . pH BUFFERS 271 We can prepare a buffer of almost any pH provided we know the pK a of the acid; and such values are easily calculated from the K a values in Table 6.5 and in most books of pHysical chemistry and Equation (6.50). We first choose a weak acid whose pK a is relatively close to the buffer pH we want. We then need to measure out accurately the volume of acid and base solutions, as dictated by Equation (6.50). Worked Example 6.12 We need to prepare a buffer of pH 9.8 by mixing solutions of ammonia and ammonium chloride solution. What volumes of each are required? Take the K a of the ammonium ion as 6 × 10 −10 . Assume the two solutions have the same concentration before mixing. Strategy: (1) We calculate the pK a of the acid. (2) We identify which component is the acid and which the base. (3) And we calculate the proportions of each according to Equation (6.50). (1) From Equation (6.50), we define the pK a as − log 10 K a . Inserting values, we obtain a pK a of 9.22. (2) The action of the buffer represents the balanced reaction, NH 4 Cl → NH 3 + HCl, so NH 4 Cl is the acid and NH 3 is the base.
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