Reactions in Aqueous Solution

Reactions in Aqueous Solution
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Solutions solute + solvent -Solute present in smaller amount -Solvent in larger amount -Can be gaseous , solid (alloy), or liquid

Aqueous Solutions Solvent is water Solute: liquid or solid, (gas?)
Properties – General Electrolytes and non electrolytes Electrolytes in water conducts electricity Nonelectrolytes does not conduct electricity Light bulb illumination of ionic compds. in water because of +ve and –ve ion formation

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One of the most important substances on Earth. Can dissolve many different substances. A polar molecule because of its unequal charge distribution. Copyright © Cengage Learning. All rights reserved

Hydration is the process in which an ion is surrounded by water molecules arranged in a specific manner. d+ d- H2O

An electrolyte is a substance that, when dissolved in water, results in a solution that can conduct electricity. A nonelectrolyte is a substance that, when dissolved, results in a solution that does not conduct electricity. nonelectrolyte weak electrolyte strong electrolyte

Ions in Aqueous Solution
Ionic Theory of Solutions Many ionic compounds dissociate into independent ions when dissolved in water These compounds that “freely” dissociate into independent ions in aqueous solution are called electrolytes. Their aqueous solutions are capable of conducting an electric current.

Ionic Theory of Solutions Not all electrolytes are ionic compounds. Some molecular compounds dissociate into ions. The resulting solution is electrically conducting, and so we say that the molecular substance is an electrolyte.

Ionic Theory of Solutions Strong and weak electrolytes. A strong electrolyte is an electrolyte that exists in solution almost entirely as ions. Most ionic solids that dissolve in water do so almost completely as ions, so they are strong electrolytes.

Ionic Theory of Solutions Strong and weak electrolytes. A weak electrolyte is an electrolyte that dissolves in water to give a relatively small percentage of ions. Double arrow represents not complete reaction, another example, CH3COOH Most soluble molecular compounds are either nonelectrolytes or weak electrolytes.

Ionization of acetic acid
CH3COOH CH3COO- (aq) + H+ (aq) A reversible reaction. The reaction can occur in both directions. Acetic acid is a weak electrolyte because its ionization in water is incomplete.

Chemical Reactions of Solutions
We must know: The nature of the reaction. The amounts of chemicals present in the solutions. Copyright © Cengage Learning. All rights reserved

Working with Solutions
Molar Concentration Molar concentration, or molarity (M), is defined as the moles of solute dissolved in one liter (cubic decimeter) of solution.

EXERCISE! A g sample of potassium phosphate is dissolved in enough water to make 1.50 L of solution. What is the molarity of the solution? 1.57 M 500.0 g is equivalent to mol K3PO4 (500.0 g / g/mol). The molarity is therefore 1.57 M (2.355 mol/1.50 L). Copyright © Cengage Learning. All rights reserved

CONCEPT CHECK! Which of the following solutions contains the greatest number of ions? 400.0 mL of 0.10 M NaCl. 300.0 mL of 0.10 M CaCl2. 200.0 mL of 0.10 M FeCl3. 800.0 mL of 0.10 M sucrose. a) contains mol of ions (0.400 L × 0.10 M × 2). b) contains mol of ions (0.300 L × 0.10 M × 3). c) contains mol of ions (0.200 L × 0.10 M × 4). d) does not contain any ions because sucrose does not break up into ions. Therefore, letter b) is correct. Copyright © Cengage Learning. All rights reserved

Dilution is the procedure for preparing a less concentrated solution from a more concentrated solution. Dilution Add Solvent Moles of solute before dilution (i) after dilution (f) = MiVi MfVf =

EXERCISE! What is the minimum volume of a 2.00 M NaOH solution needed to make mL of a M NaOH solution? 60.0 mL The minimum volume needed is 60.0 mL. M1V1 = M2V2 (2.00 M)(V1) = (0.800 M)(150.0 mL) Copyright © Cengage Learning. All rights reserved

Solution Stoichiometry
The concentration of a solution is the amount of solute present in a given quantity of solvent or solution. M = molarity = moles of solute liters of solution What mass of KI is required to make 500. mL of a 2.80 M KI solution? M KI M KI volume of KI solution moles KI grams KI 1 L 1000 mL x 2.80 mol KI 1 L soln x 166 g KI 1 mol KI x 500. mL = 232 g KI

How would you prepare 60.0 mL of 0.200 M
HNO3 from a stock solution of 4.00 M HNO3? MiVi = MfVf Mi = 4.00 Mf = 0.200 Vf = 0.06 L Vi = ? L Vi = MfVf Mi = 0.200 x 0.06 4.00 = L = 3 mL 3 mL of acid + 57 mL of water = 60 mL of solution

Types of Chemical Reactions
Precipitation Reactions A precipitation reaction occurs in aqueous solution because one product is insoluble. A precipitate is an insoluble solid compound formed during a chemical reaction in solution. For example, the reaction of sodium chloride with silver nitrate forms AgCl(s), an insoluble precipitate.

Precipitation Reactions
Precipitate – insoluble solid that separates from solution PbI2 precipitate Pb(NO3)2 (aq) + 2NaI (aq) PbI2 (s) + 2NaNO3 (aq) molecular equation Pb2+ (aq) + 2NO3- (aq) + 2Na+ (aq) + 2I- (aq) PbI2 (s) + 2Na+ (aq) + 2NO3- (aq) ionic equation Pb2+ (aq) + 2I- (aq) PbI2 (s) net ionic equation Na+ and NO3- are spectator ions

Precipitation of Lead Iodide
Pb2+ + 2I PbI2 (s) PbI2

Molecular and Ionic Equations A molecular equation is one in which the reactants and products are written as if they were molecules, even though they may actually exist in solution as ions. Pb(NO3)2 (aq) + 2NaI (aq) → PbI2 (s) + 2NaNO3 (aq)

Molecular and Ionic Equations An ionic equation, however, represents strong electrolytes as separate independent ions. This is a more accurate representation of the way electrolytes behave in solution. Pb2+ (aq) + 2NO3- (aq) + 2Na+ (aq) + 2I- (aq) → PbI2 (s) + 2Na+ (aq) + 2NO3- (aq)

Molecular and Ionic Equations Complete and net ionic equations. A net ionic equation is a chemical equation from which the spectator ions have been removed. Pb2+(aq) + 2I-(aq) → PbI2 (s) A spectator ion is an ion in an ionic equation that does not take part in the reaction. 2 Na+(aq)+ 2 NO3-(aq) → 2 Na+ (aq)+ 2 NO3- (aq)

Solubility is the maximum amount of solute that will dissolve in a given quantity of solvent at a specific temperature. (insoluble) (soluble)

CONCEPT CHECK! Write the correct formula equation, complete ionic equation, and net ionic equation for the reaction between cobalt(II) chloride and sodium hydroxide. Formula Equation: CoCl2(aq) + 2NaOH(aq) Co(OH)2(s) + 2NaCl(aq) Complete Ionic Equation: Co2+(aq) + 2Cl-(aq) + 2Na+(aq) + 2OH-(aq) Co(OH)2(s) + 2Na+(aq) + 2Cl-(aq) Net Ionic Equation: Co2+(aq) + 2Cl-(aq) Co(OH)2(s) Copyright © Cengage Learning. All rights reserved

Solving Stoichiometry Problems for Reactions in Solution
Identify the species present in the combined solution, and determine what reaction occurs. Write the balanced net ionic equation for the reaction. Calculate the moles of reactants. Determine which reactant is limiting. Calculate the moles of product(s), as required. Convert to grams or other units, as required. Copyright © Cengage Learning. All rights reserved

CONCEPT CHECK! (Part I) 10.0 mL of a 0.30 M sodium phosphate solution reacts with 20.0 mL of a 0.20 M lead(II) nitrate solution (assume no volume change). What precipitate will form? lead(II) phosphate, Pb3(PO4)2 What mass of precipitate will form? 1.1 g Pb3(PO4)2 The balanced molecular equation is: 2Na3PO4(aq) + 3Pb(NO3)2(aq) → 6NaNO3(aq) + Pb3(PO4)2(s). mol Na3PO4 present to start and mol Pb(NO3)2 present to start. Pb(NO3)2 is the limiting reactant, therefore mol of Pb3(PO4)2 is produced. Since the molar mass of Pb3(PO4)2 is g/mol, 1.1 g of Pb3(PO4)2 will form. Copyright © Cengage Learning. All rights reserved

Let’s Think About It Where are we going? To find the mass of solid Pb3(PO4)2 formed. How do we get there? What are the ions present in the combined solution? What is the balanced net ionic equation for the reaction? What are the moles of reactants present in the solution? Which reactant is limiting? What moles of Pb3(PO4)2 will be formed? What mass of Pb3(PO4)2 will be formed? Copyright © Cengage Learning. All rights reserved

CONCEPT CHECK! (Part II) 10.0 mL of a 0.30 M sodium phosphate solution reacts with 20.0 mL of a 0.20 M lead(II) nitrate solution (assume no volume change). What is the concentration of nitrate ions left in solution after the reaction is complete? 0.27 M The concentration of nitrate ions left in solution after the reaction is complete is 0.27 M. Nitrate ions are spectator ions and do not participate directly in the chemical reaction. Since there were mol of Pb(NO3)2 present to start, then mol of nitrate ions are present. The total volume in solution is 30.0 mL. Therefore the concentration of nitrate ions = mol / L = 0.27 M. Copyright © Cengage Learning. All rights reserved

Let’s Think About It Where are we going? To find the concentration of nitrate ions left in solution after the reaction is complete. How do we get there? What are the moles of nitrate ions present in the combined solution? What is the total volume of the combined solution? Copyright © Cengage Learning. All rights reserved

CONCEPT CHECK! (Part III) 10.0 mL of a 0.30 M sodium phosphate solution reacts with 20.0 mL of a 0.20 M lead(II) nitrate solution (assume no volume change). What is the concentration of phosphate ions left in solution after the reaction is complete? 0.011 M The concentration of phosphate ions left in solution after the reaction is complete is M. Phosphate ions directly participate in the chemical reaction to make the precipitate. There were mol of Na3PO4 present to start, therefore there was mol of phosphate ions present to start mol of phosphate ions were used up in the chemical reaction, therefore mol of phosphate ions is leftover ( – mol). The total volume in solution is 30.0 mL. Therefore the concentration of phosphate ions = mol / L = M. Copyright © Cengage Learning. All rights reserved

Let’s Think About It Where are we going? To find the concentration of phosphate ions left in solution after the reaction is complete. How do we get there? What are the moles of phosphate ions present in the solution at the start of the reaction? How many moles of phosphate ions were used up in the reaction to make the solid Pb3(PO4)2? How many moles of phosphate ions are left over after the reaction is complete? What is the total volume of the combined solution? Copyright © Cengage Learning. All rights reserved

Acid–Base Reactions (Brønsted–Lowry)
Acid—proton donor Base—proton acceptor For a strong acid and base reaction: H+(aq) + OH–(aq) H2O(l) Copyright © Cengage Learning. All rights reserved

Performing Calculations for Acid–Base Reactions
List the species present in the combined solution before any reaction occurs, and decide what reaction will occur. Write the balanced net ionic equation for this reaction. Calculate moles of reactants. Determine the limiting reactant, where appropriate. Calculate the moles of the required reactant or product. Convert to grams or volume (of solution), as required. Copyright © Cengage Learning. All rights reserved

Titrations In a titration, a solution of accurately known concentration is added gradually to another solution of unknown concentration until the chemical reaction between the two solutions is complete. Equivalence point – the point at which the reaction is complete Indicator – substance that changes color at (or near) the equivalence point Slowly add base to unknown acid UNTIL the indicator changes color

CONCEPT CHECK! For the titration of sulfuric acid (H2SO4) with sodium hydroxide (NaOH), how many moles of sodium hydroxide would be required to react with 1.00 L of M sulfuric acid to reach the endpoint? 1.00 mol NaOH The balanced equation is: H2SO4 + 2NaOH → 2H2O + Na2SO moles of sulfuric acid is present to start. Due to the 1:2 ratio in the equation, 1.00 mol of NaOH would be required to exactly react with the sulfuric acid. 1.00 mol of sodium hydroxide would be required. Copyright © Cengage Learning. All rights reserved

Identify each of the following species as a Brønsted acid, base, or both. (a) HI, (b) CH3COO-, (c) H2PO4- HI (aq) H+ (aq) + I- (aq) Brønsted acid CH3COO- (aq) + H+ (aq) CH3COOH (aq) Brønsted base H2PO4- (aq) H+ (aq) + HPO42- (aq) Brønsted acid H2PO4- (aq) + H+ (aq) H3PO4 (aq) Brønsted base

Acid-Base Reactions Strong and Weak Acids and Bases A strong acid is an acid that ionizes completely in water; it is a strong electrolyte.

Acid-Base Reactions Strong and Weak Acids and Bases A weak acid is an acid that only partially ionizes in water; it is a weak electrolyte. The hydrogen cyanide molecule, HCN, reacts with water to produce a small percentage of ions in solution.

Acid-Base Reactions Strong and Weak Acids and Bases A strong base is a base that is present entirely as ions, one of which is OH-; it is a strong electrolyte. The hydroxides of Group IA and IIA elements, except for beryllium hydroxide, are strong bases.

Acid-Base Reactions Strong and Weak Acids and Bases A weak base is a base that is only partially ionized in water; it is a weak electrolyte. Ammonia, NH3, is an example.

Table 4.3

Neutralization Reaction
acid + base salt + water HCl (aq) + NaOH (aq) NaCl (aq) + H2O H+ + Cl- + Na+ + OH Na+ + Cl- + H2O H+ + OH H2O

Acid-Base Reactions Neutralization Reactions Canceling the spectator ions results in the net ionic equation. Note the proton transfer. H+

What volume of a 1.420 M NaOH solution is
Required to titrate mL of a 4.50 M H2SO4 solution? WRITE THE CHEMICAL EQUATION! H2SO4 + 2NaOH H2O + Na2SO4 M acid rx coef. M base volume acid moles acid moles base volume base 4.50 mol H2SO4 1000 mL soln x 2 mol NaOH 1 mol H2SO4 x 1000 ml soln 1.420 mol NaOH x 25.00 mL = 158 mL

Titration prob. (cont.) Moles of acid = (4.5 moles/1000 mL) x 25 mL= moles 1mol of acid equivalent to 2 moles of NaOH, (2 mols of NaOH/1 mol acid) x moles of acid = moles of NaOH 0.225moles of base x1000 mL base/1.42 moles = mL

Quantitative Analysis
Volumetric Analysis Consider the reaction of sulfuric acid, H2SO4, with sodium hydroxide, NaOH: Suppose a beaker contains 35.0 mL of M H2SO4. How many milliliters of M NaOH must be added to completely react with the sulfuric acid?

Quantitative Analysis
Volumetric Analysis First we must convert the L (35.0 mL) to moles of H2SO4 (using the molarity of the H2SO4). Then, convert to moles of NaOH (from the balanced chemical equation). Finally, convert to volume of NaOH solution (using the molarity of NaOH).

Oxidation-Reduction Reactions Oxidation-reduction reactions involve the transfer of electrons from one species to another. Oxidation is defined as the loss of electrons. Reduction is defined as the gain of electrons. Oxidation and reduction always occur simultaneously.

Reaction of Sodium and Chlorine
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Rules for Assigning Oxidation States
Oxidation state of an atom in an element = 0 Oxidation state of monatomic ion = charge of the ion Oxygen = -2 in covalent compounds (except in peroxides where it = -1) Hydrogen = +1 in covalent compounds Fluorine = -1 in compounds Sum of oxidation states = 0 in compounds Sum of oxidation states = charge of the ion in ions Copyright © Cengage Learning. All rights reserved

EXERCISE! Find the oxidation states for each of the elements in each of the following compounds: K2Cr2O7 CO32- MnO2 PCl5 SF4 K = +1; Cr = +6; O = –2 C = +4; O = –2 Mn = +4; O = –2 P = +5; Cl = –1 S = +4; F = –1 K2Cr2O7; K = +1; Cr = +6; O = -2 CO32-; C = +4; O = -2 MnO2; Mn = +4; O = -2 PCl5; P = +5; Cl = -1 SF4; S = +4; F = -1 Copyright © Cengage Learning. All rights reserved

Redox Characteristics
Transfer of electrons Transfer may occur to form ions Oxidation – increase in oxidation state (loss of electrons); reducing agent Reduction – decrease in oxidation state (gain of electrons); oxidizing agent Copyright © Cengage Learning. All rights reserved

CONCEPT CHECK! Which of the following are oxidation-reduction reactions? Identify the oxidizing agent and the reducing agent. Zn(s) + 2HCl(aq) ZnCl2(aq) + H2(g) Cr2O72-(aq) + 2OH-(aq) CrO42-(aq) + H2O(l) 2CuCl(aq) CuCl2(aq) + Cu(s) Zn – reducing agent; HCl – oxidizing agent c) CuCl acts as the reducing and oxidizing agent Copyright © Cengage Learning. All rights reserved

Balancing Oxidation–Reduction Reactions by Oxidation States
Write the unbalanced equation. Determine the oxidation states of all atoms in the reactants and products. Show electrons gained and lost using “tie lines.” Use coefficients to equalize the electrons gained and lost. Balance the rest of the equation by inspection. Add appropriate states. Copyright © Cengage Learning. All rights reserved

1. What is the unbalanced equation?
Zn(s) + HCl(aq) Zn2+(aq) + Cl–(aq) + H2(g) Copyright © Cengage Learning. All rights reserved

3. How are electrons gained and lost?
1 e– gained (each atom) Zn(s) + HCl(aq) Zn2+(aq) + Cl–(aq) + H2(g) – – 2 e– lost The oxidation state of chlorine remains unchanged. Copyright © Cengage Learning. All rights reserved

4. What coefficients are needed to equalize the electrons gained and lost? 1 e– gained (each atom) × 2 Zn(s) + HCl(aq) Zn2+(aq) + Cl–(aq) + H2(g) – – 2 e– lost Zn(s) + 2HCl(aq) Zn2+(aq) + Cl–(aq) + H2(g) Copyright © Cengage Learning. All rights reserved

Zn (s) + CuSO4 (aq) ZnSO4 (aq) + Cu (s)
Zn Zn2+ + 2e- Zn is oxidized Zn is the reducing agent Cu2+ + 2e Cu Cu2+ is reduced Cu2+ is the oxidizing agent Copper wire reacts with silver nitrate to form silver metal. What is the oxidizing agent in the reaction? Cu (s) + 2AgNO3 (aq) Cu(NO3)2 (aq) + 2Ag (s) Cu Cu2+ + 2e- Ag+ + 1e Ag Ag+ is reduced Ag+ is the oxidizing agent

Oxidation-Reduction Reactions
(electron transfer reactions) 2Mg Mg2+ + 4e- Oxidation half-reaction (lose e-) O2 + 4e O2- Reduction half-reaction (gain e-) 2Mg + O2 + 4e Mg2+ + 2O2- + 4e- 2Mg + O MgO

Oxidation-Reduction Reactions The reaction of an iron nail with a solution of copper(II) sulfate, CuSO4, is an oxidation- reduction reaction The molecular equation for this reaction is:

Oxidation-Reduction Reactions The net ionic equation shows the reaction of iron metal with Cu2+(aq) to produce iron(II) ion and copper metal. Loss of 2 e-1 oxidation Gain of 2 e-1 reduction

Oxidation-Reduction Reactions Describing Oxidation-Reduction Reactions Look again at the reaction of iron with copper(II) sulfate. We can write this reaction in terms of two half-reactions.

Oxidation-Reduction Reactions Describing Oxidation-Reduction Reactions A half-reaction is one of the two parts of an oxidation-reduction reaction. One involves the loss of electrons (oxidation) and the other involves the gain of electrons (reduction). oxidation half-reaction reduction half-reaction

Oxidation-Reduction Reactions Describing Oxidation-Reduction Reactions An oxidizing agent is a species that oxidizes another species; it is itself reduced. A reducing agent is a species that reduces another species; it is itself oxidized. Loss of 2 e- oxidation reducing agent oxidizing agent Gain of 2 e- reduction

Oxidation-Reduction Reactions Oxidation Numbers The concept of oxidation numbers is a simple way of keeping track of electrons in a reaction. The oxidation number (or oxidation state) of an atom in a substance is the actual charge of the atom if it exists as a monatomic ion. Alternatively, it is hypothetical charge assigned to the atom in the substance by simple rules.

Oxidation-Reduction Reactions Oxidation Number Rules Rule Applies to Statement 1 Elements The oxidation number of an atom in an element is zero. 2 Monatomic ions The oxidation number of an atom in a monatomic ion equals the charge of the ion. 3 Oxygen The oxidation number of oxygen is –2 in most of its compounds. (An exception is O in H2O2 and other peroxides, where the oxidation number is –1.) In compounds or molecules,

Oxidation-Reduction Reactions Oxidation Number Rules Rule Applies to Statement 4 Hydrogen The oxidation number of hydrogen is +1 except when it is bonded to metals in binary compounds, where it is -1, eg. LiH 5 Halogens Fluorine is –1 in all its compounds. Each of the other halogens is –1 in binary compounds unless the other element is oxygen. 6 Compounds and ions The sum of the oxidation numbers of the atoms in a compound is zero. The sum in a polyatomic ion equals the charge on the ion.

Oxidation number (Do ex.4.5)
The charge the atom would have in a molecule (or an ionic compound) if electrons were completely transferred. Free elements (uncombined state) have an oxidation number of zero. Na, Be, K, Pb, H2, O2, P4 = 0 In monatomic ions, the oxidation number is equal to the charge on the ion. Li+, Li = +1; Fe3+, Fe = +3; O2-, O = -2 The oxidation number of oxygen is usually –2. In H2O2 and O22- it is –1.

Types of Oxidation-Reduction Reactions
Combination Reaction A + B C +3 -1 2Al + 3Br AlBr3 Al lost electrons – oxidized Br gained electrons - reduced

Decomposition Reaction C A + B +1 +5 -2 +1 -1 2KClO KCl + 3O2

Combustion Reaction A + O B +4 -2 S + O SO2 +2 -2 2Mg + O MgO

Displacement Reaction A + BC AC + B +1 +2 Sr + 2H2O Sr(OH)2 + H2 Hydrogen Displacement +4 +2 TiCl4 + 2Mg Ti + 2MgCl2 Metal Displacement -1 -1 Cl2 + 2KBr KCl + Br2 Halogen Displacement

The Activity Series for Halogens
F2 > Cl2 > Br2 > I2 Halogen Displacement Reaction -1 -1 Cl2 + 2KBr KCl + Br2 I2 + 2KBr KI + Br2

The Activity Series for Metals
Hydrogen Displacement Reaction M + BC AC + B M is metal BC is acid or H2O B is H2 Ca + 2H2O Ca(OH)2 + H2 Pb + 2H2O Pb(OH)2 + H2

Oxidation-Reduction Reactions Balancing Simple Oxidation-Reduction Reactions Since the number of electrons lost in the oxidation half-reaction must equal the number gained in the reduction half-reaction, oxidation half-reaction reduction half-reaction we must double the reaction involving the reduction of the silver.

Oxidation-Reduction Reactions Balancing Simple Oxidation-Reduction Reactions Adding the two half-reactions together, the electrons cancel, oxidation half-reaction reduction half-reaction which yields the balanced oxidation-reduction reaction.

Chemical Reactions Summary
Reactions often involve ions in aqueous solution. Many of these compounds are electrolytes. We can represent these reactions as molecular equations, complete ionic equations (with strong electrolytes represented as ions), or net ionic equations (where spectator ions have been canceled). Most reactions are either precipitation reactions, acid-base reactions, or oxidation-reduction reactions. Acid-base reactions are proton-transfer reactions. Oxidation-reduction reactions involve a transfer of electrons from one species to another.

Chemical Reactions Summary
Oxidation-reduction reactions usually fall into the following categories: combination reactions, decomposition reactions, displacement reactions, and combustion reactions. Molarity is defined as the number of moles of solute per liter of solution. Knowing the molarity allows you to calculate the amount of solute in a given volume of solution. Quantitative analysis involves the determination of the amount of a species in a material.

Operational Skills Using solubility rules.
Writing net ionic equations. Deciding whether precipitation occurs. Classifying acids and bases as weak or strong. Writing an equation for a neutralization. Writing an equation for a reaction with gas formation. Assigning oxidation numbers. Balancing simple oxidation-reduction reactions. Calculating molarity from mass and volume. Using molarity as a conversion factor. Diluting a solution. Determining the amount of a substance by gravimetric analysis. Calculating the volume of reactant solution needed. Calculating the quantity of a substance by titration.

Reactions in Aqueous Solution

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