Ionic Product of Water

9 Key excerpts on "Ionic Product of Water"

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  eBook - PDF

  Foundations of College Chemistry

  • Morris Hein, Susan Arena, Cary Willard(Authors)
  • 2016(Publication Date)
  • Wiley

    (Publisher)

  We’ve seen that water ionizes to a slight degree. This ionization is represented by these equilibrium equations: H 2 O + H 2 O ⥫  ⥬ H 3 O + + OH - (1) H 2 O ⥫  ⥬ H + + OH - (2) Equation 1 is the more accurate representation of the equilibrium because free protons (H + ) do not exist in water. Equation 2 is a simplified and often-used representation of the water equilibrium. The actual concentration of H + produced in pure water is minute and amounts to only 1.00 × 10 -7 mol L at 25°C. In pure water, [H + ] = [OH - ] = 1.00 × 10 -7 mol L since both ions are produced in equal molar amounts, as shown in equation 2. The H 2 O ⥫  ⥬ H + + OH - equilibrium exists in water and in all water solutions. A special equilibrium constant called the ion product constant for water, K w , applies to this equilibrium. The constant K w is defined as the product of the H + ion concentration and the OH - ion concentration, each in moles per liter: K w = [H + ][OH - ] The numerical value of K w is 1.00 × 10 -14 for pure water at 25°C, K w = [H + ][OH - ] = (1.00 × 10 -7 )(1.00 × 10 -7 ) = 1.00 × 10 -14 The value of K w for all water solutions at 25°C is the constant 1.00 × 10 -14 . It is impor- tant to realize that as the concentration of one of these ions, H + or OH - , increases, the other decreases. However, the product of [H + ] and [OH - ] always equals 1.00 × 10 -14 . This relationship can be seen in the examples shown in TABLE 16.1. If the concentra- tion of one ion is known, the concentration of the other can be calculated from the K w expression. LEARNING OBJECTIVE KEY TERM ion product constant for water, K w EXAMPLE 16.6 Calculate K eq for the following reaction based on concentrations of PCl 5 = 0.030 mol L, PCl 3 = 0.97 mol L, and Cl 2 = 0.97 mol L at 300°C.

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  Basic Analytical Chemistry

  • L. Pataki, E. Zapp, R. Belcher, D Betteridge, L Meites(Authors)
  • 2013(Publication Date)
  • Pergamon

    (Publisher)

  (32) Applying the law of mass action to this equilibrium [ H 3 0 + ] [ O H -] = g [H 2 0P (concentrations rather than activities are used, for simplicity). As the concentration of water in the solution is essentially invariant, this can be simplified to: K w = [H+][OH-] = 1-0 X IO 14 at 25°C (33) where K w is the Ionic Product of Water. In pure water, dissociation according to equation (32) yields hydro-gen ions* and hydroxide ions in equal amounts, hence: [H+][OH-] = [H+p = [OH-]» = K w . or [H+] = [OH] = 1-0 X IO 7 at 25°C. (34) Solutions in which the hydrogen ion and hydroxide ion concentrations are equal are said to be neutral. It follows from this equation that one mole in ten million dm 3 of water is dissociated to ions, that is, only one molecule per 556 million molecules of water is dissociated. Even though their concentration in pure water is extremely small, the pro-ducts of the dissociation, the hydrogen and hydroxide ions, play a very important role in the reactions occurring in aqueous solutions. It also follows that hydrogen ions and hydroxide ions are present simultaneously in water, or in any aqueous solution. The concentra-tion of one of the ions can be varied arbitrarily, but the concentration of the other ion will always assume a value to make the product of their concentrations 1·0χ10~ 1 4 . When an acid is added to an aqueous * Although it has been shown that hydrogen ions are highly solvated in water (§ 1.2.1), for simplicity they are given the symbol H + in the subsequent discussion. 3 20 BASIC ANALYTICAL CHEMISTRY system, the hydrogen ion concentration increases, and the concen-tration of hydroxide ions decreases simultaneously. When a base is added, an opposite process will take place. In the reaction of an acid with a base the hydrogen and hydroxide ions combine to give water until the product of the ionic concentrations reaches the value of K w .

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  Chemistry

  The Molecular Nature of Matter

  • Neil D. Jespersen, Alison Hyslop(Authors)
  • 2014(Publication Date)
  • Wiley

    (Publisher)

  Often, for convenience, we omit the water molecule that carries the hydrogen ion and write H + in place of H 3 O + . The equilibrium equation for the autoionization of water then simplifies to H 2 O m H + + OH - The equation for K w based on this is likewise simplified. 3 H + 4 3 OH - 4 = K w (16.2) In pure water, the concentrations of H + and OH - produced by the autoionization are equal because the ions are formed in equal numbers. It’s been found that the concentra- tions have the following values at 25 °C: 3 H + 4 = 3 OH - 4 = 1.0 Ž 10 -7 mol L -1 Therefore, at 25 °C, K w = (1.0 Ž 10 -7 ) Ž (1.0 Ž 10 -7 ) K w = 1.0 Ž 10 -14 (16.3) As with other equilibrium constants, the value of K w varies with tem- perature (see Table 16.1). Unless stated otherwise, we will deal with sys- tems at 25 °C. Effect of Solutes on [H + ] and [OH - ] Water’s autoionization takes place in any aqueous solution, and because of the effects of other solutes, the molar concentrations of H + and OH - may not be equal. Nevertheless, their product, K w , is always the same. Although Equations 16.1–16.3 were derived for pure water, they also apply to dilute aqueous solutions. The significance of this must be empha- sized. In any aqueous solution, the product of [H + ] and [OH - ] equals K w , although the two molar concentrations may not actually equal each other. Criteria for Acidic, Basic, and Neutral Solutions One of the consequences of the autoionization of water is that in any aqueous solution, there are always both H 3 O + and OH - ions, regardless of what solutes are present. This means that in a solution of the acid HCl there is some OH - , and in a solution of the base NaOH, there is some H 3 O + . We call a solution acidic or basic depending on which ion has the largest concentration. A neutral solution is one in which the molar concentrations of H 3 O + and OH - are equal.

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  Introduction to General, Organic, and Biochemistry

  • Morris Hein, Scott Pattison, Susan Arena, Leo R. Best(Authors)
  • 2014(Publication Date)
  • Wiley

    (Publisher)

  In pure water, [H + ] = [OH - ] = 1.00 * 10 -7 mol > L since both ions are produced in equal molar amounts, as shown in equation 2. The H 2 O m H + + OH - equilibrium exists in water and in all water solutions. A spe- cial equilibrium constant called the ion product constant for water, K w , applies to this equilibrium. The constant K w is defined as the product of the H + ion concentration and the OH - ion concentration, each in moles per liter: K w = [H + ][OH - ] The numerical value of K w is 1.00 * 10 -14 for pure water at 25°C, K w = [H + ][OH - ] = (1.00 * 10 -7 )(1.00 * 10 -7 ) = 1.00 * 10 -14 The value of K w for all water solutions at 25°C is the constant 1.00 * 10 -14 . It is important to realize that as the concentration of one of these ions, H + or OH - , increases, the other decreases. However, the product of [H + ] and [OH - ] always equals 1.00 * 10 -14 . This relationship can be seen in the examples shown in Table 16.1. If the concentration of one ion is known, the concentration of the other can be calculated from the K w expression. K w = [H + ][OH - ] [H + ] = K w [OH - ] [OH - ] = K w [H + ] TABLE 16.1 Relationship of H + and OH - Concentrations in Water Solutions [H  ] [OH  ] K w pH pOH 1.00 * 10 -2 1.00 * 10 -12 1.00 * 10 -14 2.00 12.00 1.00 * 10 -4 1.00 * 10 -10 1.00 * 10 -14 4.00 10.00 2.00 * 10 -6 5.00 * 10 -9 1.00 * 10 -14 5.70 8.30 1.00 * 10 -7 1.00 * 10 -7 1.00 * 10 -14 7.00 7.00 1.00 * 10 -9 1.00 * 10 -5 1.00 * 10 -14 9.00 5.00 E X A M P L E 1 6 . 7 What is the concentration of (a) H + and (b) OH - in a 0.001 M HCl solution? Remember that HCl is 100% ionized. SOLUTION (a) Since all the HCl is ionized, H + = 0.001 mol > L, or 1 * 10 -3 mol > L: HCl ¡ H + + Cl - 0.001 M 0.001 M [H + ] = 1 * 10 -3 mol > L (b) To calculate the [OH - ] in this solution, use the following equation and substitute the values for K w and [H + ]: [OH - ] = K w [H + ] [OH - ] = 1.00 * 10 -14 1 * 10 -3 = 1 * 10 -11 mol > L

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  Chemistry

  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).

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  Fundamentals of Analytical Chemistry

  • Douglas Skoog, Donald West, F. Holler, Stanley Crouch, Douglas Skoog(Authors)
  • 2021(Publication Date)
  • Cengage Learning EMEA

    (Publisher)

  7B Chemical Equilibrium 171 At 25°C, 3 H 3 O 1 4 5 3 OH 2 4 5 !1.00 3 10 214 5 1.00 3 10 27 M At 100°C, from Table 7-3, 3 H 3 O 1 4 5 3 OH 2 4 5 !49 3 10 214 5 7.0 3 10 27 M EXAMPLE 7-2 Calculate the hydronium and hydroxide ion concentrations and the pH and pOH of 0.200 M aqueous NaOH at 25°C. Solution Sodium hydroxide is a strong electrolyte, and its contribution to the hydroxide ion concentration in this solution is 0.200 mol/L. As in Example 7-1, hydroxide ions and hydronium ions are formed in equal amounts from the dissociation of water. There- fore, write 3 OH 2 4 5 0.200 1 3 H 3 O 1 4 where 3 H 3 O 1 4 is equal to the hydroxide ion concentration from the dissociation of water. The concentration of OH 2 from the water is insignificant, however, compared with 0.200, so we can write 3 OH 2 4 ^ 0.200 pOH 5 2log 0.200 5 0.699 Equation 7-11 is then used to calculate the hydronium ion concentration: 3 H 3 O 1 4 5 K w 3 OH 2 4 5 1.00 3 10 214 0.200 5 5.00 3 10 214 M pH 5 2log 5.00 3 10 214 5 13.301 Note that the approximation 3 OH 2 4 5 0.200 1 5.00 3 10 214 ^ 0.200 M causes no significant error in our answer. 7B-5 Using Solubility-Product Constants Most, but not all, sparingly soluble salts are essentially completely dissociated in satu- rated aqueous solution. For example, when an excess of barium iodate is equilibrated with water, the dissociation process is adequately described by the equation Ba 1 IO 3 2 2 1 s 2 m Ba 21 1 aq 2 1 2IO 2 3 1 aq 2 When we say that a sparingly soluble salt is completely dissociated, we do not imply that all of the salt dissolves. What we mean is that the very small amount that does go into solution dissociates completely. ❯ Copyright 2022 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).

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  Drinking Water Treatment

  An Introduction

  • Eckhard Worch(Author)
  • 2019(Publication Date)
  • De Gruyter

    (Publisher)

  The fractions of the other species are available in an analogous manner. In the same way, the speciation of acids with more than two protons can also be calculated. Examples are shown in Figures 1.4, 1.5, and 1.8 (Chapter 1, Section 1.2.5). For a base B, we can write: B + H 2 O 󴀕󴀬 BH + + OH − (2.33) where BH + stands for the protonated base. According to the conventions mentioned in Section 2.3.1 (inclusion of c ( H 2 O ) in the constant), the law of mass action related to Equation (2.33) reads: K ∗ b = a ( BH + ) a ( OH − ) a ( B ) (2.34) or K b = c ( BH + ) c ( OH − ) c ( B ) (2.35) where K ∗ b is the thermodynamic basicity constant and K b is the conditional basicity constant. Analogous to the acidity constant, a logarithmic form of the basicity con-stant can be defined by: p K ∗ b = − log K ∗ b (2.36) 2.3 Chemical equilibria | 33 Water itself dissociates to a small extent into protons and hydroxide ions: H 2 O 󴀕󴀬 H + + OH − (2.37) The law of mass action for the water dissociation is given by: K ∗ w = a ( H + ) a ( OH − ) (2.38) where K ∗ w is the dissociation constant of water, which has a value of 1 × 10 − 14 mol 2 / L 2 at 25 °C. The constant K ∗ w , also referred to as ion product of water, is frequently given in logarithmic form as: p K ∗ w = − log K ∗ w (2.39) If the activities (or, simplifying, the concentrations) of the protons and the hydroxide ions are described by the parameters pH and pOH according to: pH = − log a ( H + ) ≈ − log c ( H + ) (2.40) and pOH = − log a ( OH − ) ≈ − log c ( OH − ) (2.41) then Equation (2.38) can also be written in the form: p K ∗ w = pH + pOH (2.42) with p K ∗ w = 14 at 25 °C.

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  The Chemistry Companion

  • Anthony C. Fischer-Cripps(Author)
  • 2011(Publication Date)
  • CRC Press

    (Publisher)

  9.2 Ionisation of Water Experiments indicate that pure water but it does conduct, indicating the 9. Ionic Equilibrium electrolyte . Water dissociates into io n     OH H O H 2 The degree of dissociation is very s m and OH  ions is also very small. The concentration of water, exp r established from its molecular weight 16 m.w. O H 2   T h m b u th 18  One litre of water has a mass of 1000 litre of water is: mol e 6 . 55 18 1000   n That is, water has a concentration of 5 Since the concentration of the H + exceedingly small, the concentration the ion p roduct K w is forme d : w            O H H O H O H H w 2 K K Experiments show that the ion p 1.0  10 -14 mole 2 /litre 2 . The concen t 10  7 moles/litre respectively. A solution containing equal conce n concentrations are said to be neutral .       m 10 1 OH m 10 1 H 7 7         is a very poor conductor of electricity, presence of ions. Water is a weak 111 n s according to m all, and so the concentration of the H + r esse d as moles/litre, can be readily and its density: 2  h e H + ions attach themselves to other water olecules to form the hydronium ion H 3 O + , u t here we will simply write the H + ions as if ey exist on their own. g, therefore the number of moles n in a e s 5 5.6M. + and OH  ions in a litre of water is of H 2 O is virtually a constant, and so      H H p roduct of pure water at 25  C is t rations of H + and OH  ions are 1.0  The concentration of H 2 O (55.6M) is taken to be a constant and incorporated into K w . n trations of H + and OH  ions at these m ol/litre m ol/litre 9.3 H + and OH  Concentrati o Consider a 0.1M solution of HCl: 112    C H HCl The concentration of H + ions will be 0 the small amount of H + ions from th e the equilibrium between the H + and t h established such that That is, compared to neutral water, t reduced from 1  10  7 to 1  10  13 due t Consider now an aqueous solutio    1 1 OH 1 .

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  Soil and Water Chemistry

  An Integrative Approach, Second Edition

  • Michael E. Essington(Author)
  • 2015(Publication Date)
  • CRC Press

    (Publisher)

  222 Soil and Water Chemistry: An Integrative Approach throughout the normal pH range of soil solutions. Ions with moderate to high IP values (0.03 < IP < 0.1) tend to strongly polarize water and promote hydrolysis: [M(H 2 O) n ] m + = [MOH(H 2 O) n −1 ] m −1 + H + , where M is a metal cation with charge m +. This type of hydrolysis behavior is common to Al 3+ and Fe 3+ . Ions with high IP values (>0.1) promote the complete dissociation of waters of hydration, resulting in the formation of stable oxyanions: [M(H 2 O) n ] m + = [MO n ] m −2 n + 2 n H + . Indeed, metals that display a high IP do not exist as cations in aqueous solutions. The degree of hydrolysis is also pH dependent. In acidic solutions, proton activity is high and dis-sociation of a proton from a water of hydration (to essentially put more protons into solution) is not favored. In alkaline solutions, proton activity is low and hydrolysis, or the dissociation of protons from waters of hydration, is favored. The degree of hydrolysis is described by an equilibrium constant. Consider the hydrolysis of Al(H 2 O) 6 3+ : Al(H 2 O) 6 3+ → AlOH(H 2 O) 5 2+ + H + (5.25) The equilibrium constant for this reaction is (25°C and 0.101 MPa; Table 5.7): K a = + + + (AlOH(H O) )(H ) (Al(H O) ) 2 5 2 2 6 3 (5.26) and K a is termed the acid dissociation constant . The negative log of the K a (–log K a ) is designated p K a (p simply means –log).

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