Chemistry
Henderson-Hasselbalch Equation
The Henderson-Hasselbalch equation is a mathematical expression used to calculate the pH of a solution containing a weak acid and its conjugate base, or a weak base and its conjugate acid. It relates the pH of the solution to the pKa of the acid and the concentrations of the acid and its conjugate base. The equation is pH = pKa + log([A-]/[HA]), where A- is the conjugate base and HA is the weak acid.
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12 Key excerpts on "Henderson-Hasselbalch Equation"
eBook - PDF- Paul Flowers, Klaus Theopold, Richard Langley, William R. Robinson(Authors)
- 2015(Publication Date)
- Openstax(Publisher)
This equation relates the pH, the ionization constant of a weak acid, and the concentrations of the weak acid and its salt in a buffered solution. Scientists often use this expression, called the Henderson-Hasselbalch Equation, to calculate the pH of buffer solutions. It is important to note that the “x is small” assumption must be valid to use this equation. Lawrence Joseph Henderson and Karl Albert Hasselbalch Lawrence Joseph Henderson (1878–1942) was an American physician, biochemist and physiologist, to name only a few of his many pursuits. He obtained a medical degree from Harvard and then spent 2 years studying in Strasbourg, then a part of Germany, before returning to take a lecturer position at Harvard. He eventually became a professor at Harvard and worked there his entire life. He discovered that the acid-base balance in human blood is regulated by a buffer system formed by the dissolved carbon dioxide in blood. He wrote an equation in 1908 to describe the carbonic acid-carbonate buffer system in blood. Henderson was broadly knowledgeable; in addition to his important research on the physiology of blood, he also wrote on the adaptations of organisms and their fit with their environments, on sociology and on university education. He also founded the Fatigue Laboratory, at the Harvard Business School, which examined human physiology with specific focus on work in industry, exercise, and nutrition. In 1916, Karl Albert Hasselbalch (1874–1962), a Danish physician and chemist, shared authorship in a paper with Christian Bohr in 1904 that described the Bohr effect, which showed that the ability of hemoglobin in the blood to bind with oxygen was inversely related to the acidity of the blood and the concentration of carbon dioxide. The pH scale was introduced in 1909 by another Dane, Sørensen, and in 1912, Hasselbalch published measurements of the pH of blood.
eBook - PDFGeneral, Organic, and Biological Chemistry
An Integrated Approach
- Kenneth W. Raymond(Author)
- 2012(Publication Date)
- Wiley(Publisher)
262 CHAPTER 7 Acids, Bases, and Equilibrium We have seen that a buffer can be prepared from a weak acid and its conjugate base, and that the concentration of these two substances is connected to the pH of the solution and the pK a of the acid (Table 7.6). Taking a more mathematical approach to these concepts allows us to more accurately describe the makeup of a buffer. Beginning with the equation for the K a of the acid HA, HA + H 2 O N A - + H 3 O + K a = [A - ][H 3 O + ] [HA] taking the logarthim of both sides of the equation, and rear- ranging gives what is known as the Henderson–Hasselbalch equation. pH = pK a + log [A - ] [HA] The Henderson–Hasselbalch equation To see how this equation can be used, let us consider the acid HF (pK a = 3.18). HF + H 2 O N F - + H 3 O + K a = [F - ][H 3 O + ] [HF] and pH = pK a + log [F - ] [HF] • What is the pH when [F - ] = [HF]? If [F - ] = [HF], then [F - ] [HF] = 1. Inserting this and the value for the pK a of HF into the Henderson–Hasslebalch equation gives a pH of 3.18. This agrees with the prediction in Table 7.6 that if the values of the pH and the pK a for an acid are the same, then [HA] = [A - ]. pH = pK a + log [F - ] [HF] = 3.18 + log 1 = 3.18 + 0 = 3.18 • What is the pH when [HF] is two times larger than [F - ]? In mathematical terms, this relationship can be expressed as [HF] = 2[F - ] or [F - ] [HF] = 0.5. Solving the Henderson– Hasselbalch equation gives a pH of 2.88. As predicted by Table 7.6, when [HA] 7 [A - ], pH 6 pK a . pH = pK a + log [F - ] [HF] = 3.18 + log 0.5 = 3.18 + -0.30 = 2.88 • What is the pH when [HF] is half that of [F - ]? In math- ematical terms, this relationship can be expressed as 2[HF] = [F - ] or [F - ] [HF] = 2. Solving the Henderson–Hasselbalch equation gives a pH of 3.48.
eBook - ePubRapid Review of Chemistry for the Life Sciences and Engineering
With Select Applications
- Armen S. Casparian, Gergely Sirokman, Ann Omollo(Authors)
- 2021(Publication Date)
- CRC Press(Publisher)
follows: K h = [ H 3 O + ] [ NH 3 ] [ NH 4 + ] The value for K h is not given but can be found by applying the relationship K h = K w K b = 1.0 × 10 − 14 1.8 × 10 − 5 = 5.56 × 10 − 10 where K b is the value of the equilibrium base constant for NH 3 in H 2 O and is found in Table 4.1B. It is analogous to K a : NH 3 + H 2 O ⇋ NH 4 + + OH − The same strategy is then used as in Example 4.8. At equilibrium, then, [ H 3 O + ] = x [ NH 3 ] = x [ NH 4 + ] = 0.20 − x Therefore, 5.56 × 10 − 10 = (x) (x) 0.20 − x Use the same simplifying assumption as in Example 4.8, assume x ≪ 0.20 M, and solve. for x : x 2 = (0.20 M) (5.56 × 10 − 10) x = 1.1 × 10 − 5 Thus, [ H 3 O + ] = 1.1 × 10 − 5 M And, pH = − log [ H 3 O + ] = − log (1.1 × 10 − 5 M) = 4.96 The pH is less than 7.0, reflecting an acidic solution, as expected. Note: For the hydrolysis of an ion producing a basic or alkaline solution, K h = K w /K a. 4.9. Buffer Solutions and the Henderson–Hasselbalch Equation Buffer solutions are solutions that protect against large shifts in pH in the event of a shock due to the sudden addition of a strong acid or base. The pH of a buffer solution does, however, change slightly. Such a solution can be made with a combination of either of the following: A weak acid and its salt (conjugate base) A weak base and its salt (conjugate acid) Thus, acetic acid plus sodium acetate constitutes a buffer system, as does ammonium hydroxide plus ammonium chloride. To determine the pH of a buffer solution, the Henderson–Hasselbalch equation may be used. In the case of a weak acid (e.g., CH 3 COOH) and its conjugate base salt (e.g., NaCH 3 COO): pH = p K a + log [ conjugate base ] [ weak acid ] (4.8) The buffering capacity of a system refers to the amount of acid or base it can absorb before its pH changes
eBook - PDFChemistry
The Molecular Nature of Matter
- Neil D. Jespersen, Alison Hyslop(Authors)
- 2021(Publication Date)
- Wiley(Publisher)
With these as conditions, the equation is sometimes written pH = p K a + log [salt] _____ [acid] (16.20) For practice, you may wish to apply this equation to buffer problems at the end of the chapter. NOTE “Taking the negative logarithm” means we take the logarithm and change its mathematical sign if positive to negative or if negative to positive. TOOLS Henderson–Hasselbalch equation 822 CHAPTER 16 Acid–Base Equilibria in Aqueous Solutions Calculating pH Change for a Buffer Earlier we described how a buffer is able to neutralize small amounts of either strong acid or base to keep the pH from changing by large amounts. We can use this knowledge now to calculate how much the pH will change. EXAMPLE 16.7 Calculating How a Buffer Resists Changes in pH How much will the pH change if 0.020 mol of HCl is added to a buffer solution that was made by dissolving 0.12 mol of NH 3 and 0.095 mol of NH 4 Cl in 250 mL of water? Analysis: This problem will require two calculations. The first is to determine the pH of the original buffer. The second is to determine the pH of the mixture after the HCl has been added. For the second calculation, we have to determine how the acid changes the amounts of NH 3 and NH 4 + . Assembling the Tools: For the pH of the initial solution we need the chemical reaction and the K b equation for NH 3 . NH 3 + H 2 O NH 4 + + OH − K b = [NH 4 + ][OH − ] ____________ [NH 3 ] = 1.8 × 10 −5 To determine the pH of the solution after the HCl was added we need the chemical reaction for HCl in this mixture. Since HCl is a strong acid, it will react quantitatively (completely) with the conjugate base of the buffer—in this case, NH 3 . The balanced molecular equation is NH 3 + HCl ⟶ NH 4 Cl In this case it is better to write the net ionic equation as NH 3 + H + ⟶ NH 4 + We then do a limiting reactant calculation to determine how much of the NH 3 will be converted to NH 4 + . We then use the new amounts of NH 3 and NH 4 + to determine the pH.
- Frederick Bettelheim, William Brown, Mary Campbell, Shawn Farrell(Authors)
- 2019(Publication Date)
- Cengage Learning EMEA(Publisher)
HA 1 H 2 O m A 2 1 H 3 O 1 K a 5 [A 2 ][H 3 O 1 ] [HA] Taking the logarithm of this equation gives: log K a 5 log 31 [H 3 O 1 ] [A 2 ] [HA] 24 5 log [H 3 O 1 ] 1 log [A 2 ] [HA] Rearranging terms gives us a new expression, in which 2 log K a is, by def-inition, p K a and 2 log [H 3 O 1 ] is, by definition, pH. Making these substitu-tions gives the Henderson-Hasselbalch Equation. 2 log [H 3 O 1 ] 5 2 log K a 1 log [A 2 ] [HA] pH 5 p K a 1 log Henderson–Hasselbalch equation [A 2 ] [HA] Because [A 2 ] [HA] is a ratio, it doesn’t matter if units are given in terms of concentration, moles, or volumes when using the Henderson–Hasselbalch equation, as long as consistent units are used when calculating the ratio. The Henderson–Hasselbalch equation gives us a convenient way to cal-culate the pH of a buffer when the concentrations of the weak acid and its conjugate base are not equal. EXAMPLE 8.11 Buffer pH Calculation What is the pH of a phosphate buffer solution containing 1.0 mol/L of sodium dihydrogen phosphate, NaH 2 PO 4 , and 0.50 mol/L of sodium hydrogen phosphate, Na 2 HPO 4 ? STRATEGY Use the Henderson–Hasselbalch equation to determine the pH. You must know either the number of moles of both the conjugate acid and base or the concentrations of the conjugate acid or base. Divide the conjugate base by the conjugate acid, take the log of that ratio, and add it to the p K a of the conjugate acid. SOLUTION The weak acid in this problem is H 2 PO 4 2 ; its ionization produces HPO 4 2 2 . The p K a of this acid is 7.21 (from Table 8.3). Under the weak acid and its conjugate base are shown their concentrations. 1 H 2 PO 4 2 1.0 mol/L HPO 4 2 2 0.50 mol/L H 2 O 1 H 3 O 1 p K a 5 7.21 256 | Chapter 8 Acids and Bases Copyright 2020 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
eBook - ePub- Peter Kam, Ian Power(Authors)
- 2015(Publication Date)
- CRC Press(Publisher)
K) of the substance.P H SYSTEMH+ ion concentration may be measured in two ways: directly as concentrations in nanomoles per litre or indirectly as pH. pH is defined as the negative logarithm (to the base 10) of the concentration of hydrogen ions. The pH is related to the concentration of H+ as follows:pH = log101[H +]pH = log10[H +]H +=10− pHpH = p K + log base/acidTable 8.1 Relationship between pH and hydrogen ion concentrationpH Hydrogen ion concentration (nmol/L) 7.7 20 7.4 40 7.3 50 7.1 80 It is important to note that pH and hydrogen ion concentration [H+ ] are inversely related such that an increase in pH describes a decrease in [H+ ] (Table 8.1 ). However, the logarithmic scale is nonlinear and, therefore, a change of one pH unit reflects a 10-fold change in [H+ ] and equal changes in pH are not correlated with equal changes in [H+ ]. For example, a change of pH from 7.4 to 7.0 (40 nmol/L [H+ ] to 100 nmol/L [H+ ]) represents a change of 60 nmol/L [H+ ], although the same pH change of 0.4, but from 7.4 to 7.8 (40 nmol/L [H+ ] to 16 nmol/L [H+ ]), represents a change of only 24 nmol/L [H+ ].BUFFERSA buffer is a solution consisting of a weak acid and its conjugate base, which resists a change in pH when a stronger acid or base is added, thereby minimizing a change in pH. The most important buffer pair in extracellular fluid (ECF) is carbonic acid (H2 CO3 ) and bicarbonate (HYDROGEN ION BALANCEHCO 3 −). The interaction between this buffer pair forms the basis of the measurement of acid–base balance.Cellular hydrogen ion turnover can be described in terms of processes that produce or consume H+ ions in the body (Table 8.2 ). The total daily H+
eBook - PDFChemistry
The Molecular Nature of Matter
- James E. Brady, Neil D. Jespersen, Alison Hyslop(Authors)
- 2014(Publication Date)
- Wiley(Publisher)
How buffers work (Section 16.7) These reactions illustrate how the concentrations of conjugate acid and base change when a strong acid or strong base is added to a buffer. Adding H + decreases [A - ] and increases [H A]; adding OH - decreases [H A] and increases [A - ]. These reactions keep the pH of the solution from changing drastically. Henderson–Hasselbalch equation (Section 16.7) pH = pK a + log 3 A - 4 3 H A 4 Polyprotic acids ionize stepwise (Section 16.8) Polyprotic acids have more than one ionizable proton in their formulas. Each proton will ionize sequentially, and the chemical equation will have a form similar to the general equation for the ionization of a weak acid tool above. Titration curves have four parts (Section 16.9) To plot the graph of pH as a function of volume of titrant added, that is, a titration curve, there are four different parts of calcula- tion. They are (a) the starting point, (b) the part from the start to the equivalence point, (c) the equivalence point, and (d) the part after the equivalence point. How acid–base indicators work (Section 16.9) Titration indicators are weak acids that have one color for the conjugate acid and a different color for the conjugate base, and the change in color depends on the pH of the solution. | Review Questions Water, pH, and “p” Notation 16.1 When [H 3 O + ] equals 3.80 Ž 10 -5 M at 25 °C, (a) What is the value of [OH - ]? (b) Is the solution acidic, basic, or neutral? 16.2 At 37 °C, K w = 2.5 Ž 10 -14 . If an aqueous solution has [H 3 O + ] equal to 1.6 Ž 10 -7 M, is the solution acidic, basic, or neutral? 16.3 Calculate the concentration of H + in a solution in which the hydroxide ion concentration is (a) 2.0 Ž 10 -5 M (b) 4.0 Ž 10 -8 M 16.4 An aqueous solution at 25 °C with a pOH of 10.00 has a hydrogen ion concentration equal to (a) 1.00 Ž 10 -10 M (c) 10.00 M (b) 1.00 Ž 10 -4 M (d) 4.00 M = WileyPLUS, an online teaching and learning solution.
eBook - PDFChemistry
The Molecular Nature of Matter
- Neil D. Jespersen, Alison Hyslop(Authors)
- 2014(Publication Date)
- Wiley(Publisher)
How buffers work (Section 16.7) These reactions illustrate how the concentrations of conjugate acid and base change when a strong acid or strong base is added to a buffer. Adding H + decreases [A - ] and increases [H A]; adding OH - decreases [H A] and increases [A - ]. These reactions keep the pH of the solution from changing drastically. Henderson–Hasselbalch equation (Section 16.7) pH = pK a + log 3 A - 4 3 H A 4 Polyprotic acids ionize stepwise (Section 16.8) Polyprotic acids have more than one ionizable proton in their formulas. Each proton will ionize sequentially, and the chemical equation will have a form similar to the general equation for the ionization of a weak acid tool above. Titration curves have four parts (Section 16.9) To plot the graph of pH as a function of volume of titrant added, that is, a titration curve, there are four different parts of calcula- tion. They are (a) the starting point, (b) the part from the start to the equivalence point, (c) the equivalence point, and (d) the part after the equivalence point. How acid–base indicators work (Section 16.9) Titration indicators are weak acids that have one color for the conjugate acid and a different color for the conjugate base, and the change in color depends on the pH of the solution. | Review Questions Water, pH and “p” Notation 16.1 Write the chemical equation for (a) the autoionization of water and (b) the equilibrium law for K w . 16.2 How are acidic, basic, and neutral solutions in water defined (a) in terms of [H + ] and [OH - ] and (b) in terms of pH and pOH? 16.3 At 25 °C, how are the pH and pOH of a solution related to each other? 16.4 Why do chemists use pH notation instead of the concen- tration of H + ions? 16.5 Explain how acids and bases suppress the ionization of water, often called the common ion effect.
eBook - PDFChemistry
An Atoms First Approach
- Steven Zumdahl, Susan Zumdahl, Donald J. DeCoste, , Steven Zumdahl, Steven Zumdahl, Susan Zumdahl, Donald J. DeCoste(Authors)
- 2020(Publication Date)
- Cengage Learning EMEA(Publisher)
Since K a values rather than pK a values are given for the various acids, we use Equa- tion (14.1) fH 1 g 5 K a fHAg fA 2 g rather than the Henderson–Hasselbalch equation. We substitute the required [H 1 ] and K a for each acid into Equation (14.1) to calculate the ratio [HA]y[A 2 ] needed in each case. Acid fH 1 g 5 K a fHAg fA 2 g fHAg fA 2 g a. Chloroacetic 5.0 3 10 25 5 1.35 3 10 23 S fHAg fA 2 g D 3.7 3 10 22 b. Propanoic 5.0 3 10 25 5 1.3 3 10 25 S [HA] [A 2 ] D 3.8 c. Benzoic 5.0 3 10 25 5 6.4 3 10 25 S [HA] [A 2 ] D 0.78 d. Hypochlorous 5.0 3 10 25 5 3.5 3 10 28 S [HA] [A 2 ] D 1.4 3 10 3 Since [HA]y[A 2 ] for benzoic acid is closest to 1, the system of benzoic acid and its sodium salt will be the best choice among those given for buffering a solution at pH 4.30. This example demonstrates the principle that the optimal buffering system has a pK a value close to the desired pH. The pK a for benzoic acid is 4.19. 14.4 Titrations and pH Curves As we saw in Chapter 6, a titration is commonly used to determine the amount of acid or base in a solution. This process involves a solution of known concentration (the titrant) delivered from a buret into the unknown solution until the substance being ana- lyzed is just consumed. The stoichiometric (equivalence) point is often signaled by the color change of an indicator. In this section we will discuss the pH changes that occur during an acid–base titration. We will use this information later to show how an appropriate indicator can be chosen for a particular titration. The progress of an acid–base titration is often monitored by plotting the pH of the solution being analyzed as a function of the amount of titrant added. Such a plot is called a pH curve or titration curve.
eBook - PDFChemistry
Structure and Dynamics
- James N. Spencer, George M. Bodner, Lyman H. Rickard(Authors)
- 2011(Publication Date)
- Wiley(Publisher)
Three primary buffer systems maintain the pH of blood in the 510 CHAPTER 11 / ACIDS AND BASES We can start by substituting the values of the pH of the buffer solution and the pK a value for the conjugate acid into the Henderson–Hasselbalch equation. Solving for the ratio of the concentrations of the conjugate acid–base pair gives the following result. Thus if the concentration of the base is 1.3 times that of the concentration of the acid, the solution would be an effective buffer at a pH of 9.35. Suppose that 1.0 L of 0.10 M NH 3 was available. The equilibrium con- stant for the reaction between NH 3 and water is so small that the concentra- tion of NH 3 at equilibrium is essentially 0.10 M. We can calculate the concentration of the NH 4 needed to prepare a pH 9.35 buffer as follows. Because adding 0.077 mole of solid NH 4 Cl to the 0.10 M NH 3 solution would produce a negligible change in the volume of the solution, the result would be 1.0 L of a buffer with the desired pH. [NH 4 + ] = 0.077 M [0.10 M] [NH 4 + ] = 1.3 [NH 3 ] [NH 4 + ] = 1.3 NH 3 (aq) + H 2 O(l) uv NH 4 + (aq) + OH - (aq) K b = 1.8 * 10 - 5 [conjugate base] [conjugate acid] = 1.3 log a [conjugate base] [conjugate acid] b = 9.35 - 9.25 = 0.10 9.35 = 9.25 + log a [conjugate base] [conjugate acid] b pH = pK a + log a [conjugate base] [conjugate acid] b ➤ CHECKPOINT The pH of a buffered solution is 7.4. What happens to the pH if an acid is added to the buffer? What happens to the pH if a base is added to the buffer? Does the resulting pH depend on how much acid or base is added? pH meters are often calibrated with buffer solutions because the pH of these solutions is stable. [Copyright of Thermo Fisher Scientific Inc., 2010.] human body within the limits 7.36 to 7.42. One of these buffers is composed of carbonic acid, H 2 CO 3 , and its conjugate base, HCO 3 , the hydrogen carbonate ion.
eBook - PDFAnalytical Chemistry
A Toolkit for Scientists and Laboratory Technicians
- Bryan M. Ham, Aihui MaHam(Authors)
- 2024(Publication Date)
- Wiley(Publisher)
9.14.3 Natural Buffer Solutions Buffered solutions are important in many areas of analytical chemistry, biology, biochemistry, bacteriology, microbiology, and physiology. Both synthetic organic reactions are often controlled by pH and physiological reactions involving biosyn- thesis and the action of enzymes. The major intracellular buffer in living organisms is the conjugate acid–base pair of dihydrogenphosphate–monohydrogenphosphate (H 2 PO − 4 –HPO 2− 4 ), while the major extracellular buffer is the conjugate acid–base pair of carbonic acid–bicarbonate (H 2 CO 3 –HCO 3 − ). The carbonic acid–bicarbonate conjugate acid–base pair is the buffer system responsible for the maintenance of an approximate 7.4 pH in blood. 9.14.4 Calculating Buffer pH A very useful relationship for calculating the pH of buffers, or the amounts of the conjugate acid–base pair needed is derived as follows: K a = [H + ][A − ] [HA] (9.70) [H + ] = K a × [HA] [A − ] (9.71) − log[H + ] = − log K a − log [HA] [A − ] (9.72) pH = pK a − log [HA] [A − ] (9.73) pH = pK a + log [A − ] [HA] (9.74) Titrant ml of base (OH – ) Titrant ml of acid (H + ) 0 0 1 1 2 2 3 3 4 4 5 5 6 6 7 7 pH 7.0 8.0 9.0 10.0 11.0 6.0 5.0 4.0 3.0 Buffer region 100% BH + Midpoint 100% B FIGURE 9.7 Titration curve of a buffered solution. 174 ACID–BASE THEORY AND BUFFER SOLUTIONS 9.14.5 Buffer pH Calculation I What is the ratio of [HPO 4 2− ]:[H 2 PO 4 − ] required to keep an in- tracellular compartment at a pH of 7.8? pH = pK a + log [A − ] [HA] (9.75) 7.8 = − log(6.2 × 10 −8 ) + log [HPO 2− 4 ] [H 2 PO − 4 ] (9.76) 7.8 = 7.2 + log [HPO 2− 4 ] [H 2 PO − 4 ] (9.77) [HPO 2− 4 ] [H 2 PO − 4 ] = antilog(0.6) (9.78) [HPO 2− 4 ] [H 2 PO − 4 ] = 4 (9.79) From this, we see that the ratio of [HPO 4 2− ]:[H 2 PO 4 − ] needed to keep the pH at 7.8 would be 4 : 1. This is a theoretical approach to calculating the ratio of acid to conjugate base needed to pre- pare a buffer solution.
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
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