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 ELECTROLYTIC PROPERTIES

METHOD OF DISTINGUISHING BETWEEN ELECTROLYTES AND NONELECTROLYTES A pair of inert electrodes (Cu or Pt) is immersed in a beaker of water. To light the bulb, electric current must flow from one electrode to the other, thus completing the circuit. By adding NaCl (ionic compound), the bulb will glow. NaCl breaks up into Na + and Cl - ions when dissolves in water. Na + are attracted to the negative electrode. Cl - are attracted to the positive electrode. The movement sets up an electric current that is equivalent to the flow of electrons along a metal wire.

Strong Electrolyte – 100% dissociation (breaking up of compound into cations and anions NaCl (s) Na + (aq) + Cl - (aq) H2OH2O Weak Electrolyte – not completely dissociated CH 3 COOH CH 3 COO - (aq) + H + (aq) A reversible reaction. The reaction can occur in both directions.

Hydration is the process in which an ion is surrounded by water molecules arranged in a specific manner. Hydration helps to stabilize ions and prevents cations from combining with anions.   H2OH2O 

Nonelectrolyte does not conduct electricity? No cations (+) and anions (-) in solution C 6 H 12 O 6 (s) C 6 H 12 O 6 (aq) H2OH2O

Have a sour taste. - Vinegar owes its taste to acetic acid. - Citrus fruits contain citric acid. React with certain metals to produce hydrogen gas. React with carbonates and bicarbonates to produce carbon dioxide gas Cause color changes in plant dyes. 2HCl (aq) + Mg (s) MgCl 2 (aq) + H 2 (g) 2HCl (aq) + CaCO 3 (s) CaCl 2 (aq) + CO 2 (g) + H 2 O (l) Aqueous acid solutions conduct electricity. PROPERTIES ACIDS

Have a bitter taste. Feel slippery. Many soaps contain bases. Cause color changes in plant dyes. Aqueous base solutions conduct electricity. Examples: PROPERTIES OF BASES

ROLE OF WATER TO SHOW PROPERTIES OF ACIDS Anhydrous pure acid (without water) does not show acidic properties. In dry form, acids exist as neutral covalent molecules. Dry acids do not dissociate to form hydrogen ion (H + ). When a pure acid is dissolved in water, it will show the properties of acids. This is because acids will dissociate in water to form H + or hydroxonium/hydronium ion, H 3 O + which are free to move. For example: i) HCl in liquid methylbenzene (organic solvent) - does not show acidic properties. ii) HCl in water – show acidic properties

ROLE OF WATER TO SHOW PROPERTIES OF ALKALI Dry base does not show alkaline properties. A base in dry form, contains hydroxide ions (OH-) that are not free to move. Thus, the alkaline properties cannot be shown. In the presence of water, bases can dissociate in water to form hydroxide ions, OH-, which are free to move. Thus, alkaline properties are shown. For example: i) ammonia in tetrachlomethane (organic solvent) – do not show alkaline properties ii) ammonia in water – show alkaline properties

DEFINITION OF ACID AND BASE Arrhenius Brønsted- Lowry Lewis

 Arrhenius acid is a substance that produces H + (hydrogen ion) or hydronium ion (H 3 O + ) in water  Arrhenius base is a substance that produces OH - in water DEFINITION OF ACID AND BASE BY ARRHENIUS

Examples of acid:  CO 2 (g) + H 2 O (l) H 2 CO 3 (aq) H 2 CO 3 (aq) + H 2 O(l)H 3 O + (aq) + HCO 3 - (aq)  nonmetal oxides + H 2 Oacid Examples of bases:  NaOH (s) Na + (aq) + OH - (aq)  N 2 H 4 (aq) + H2O N 2 H 5 + (aq) + OH - (aq)  metal oxides + H 2 O bases Na 2 O (s) + H 2 O (l)2NaOH (aq) * Limited only to aqueous solutions

A Brønsted acid is a proton donor A Brønsted base is a proton acceptor Example: acidbaseacidbase A Brønsted acid must contain at least one ionizable proton! DEFINITION OF ACID AND BASE BY BRØNSTED-LOWRY HCl (aq) +H 2 O (l) → H 3 O + (aq) + Cl - (aq) HCl is a acid because it donates proton to H 2 O H 2 O is a base because it accepts proton from HCl

Brønsted acids and bases Conjugate acid-base pair: i) Conjugate base of a Brønsted acid - the species that remains when one proton has been removed from the acid ii) Conjugate acid - addition of a proton to a Brønsted base

Examples: HCl (aq) +H 2 O (l) H 3 O + (aq) + Cl - (aq) acid 1 base 2 acid 2 base 1 Cl - is a conjugate base of HCl and HCl is a conjugate acid of Cl - H 2 O is a base conjugate of H 3 O + and H 3 O + is a acid conjugate of H 2 O NH 3 (aq) + H 2 O (l) NH 4 + (aq) + OH - (aq) base 1 acid 2 acid 1 base 2 subscripts 1 and 2 = two conjugate acid-base pair

When a strong acid react with a strong base in Brønsted acid-base reaction, it will give a weak conjugate acid and conjugate base. Examples: HCl (aq) + H 2 O (l) H 3 O + (aq) + Cl - (aq) strong acid strong base weak conjugate weak conjugate acid base NH 3 (aq) + H 2 O (l) NH 4 + (aq) + OH - (aq) Weak base weak acid strong conjugate strong conjugate acid base H 2 O can function as acid or base which called amphoteric Amphoteric or amphiprotic substance is one that can react as either an acid or base

Identify each of the following species as a Brønsted acid, base, or both. (a) HI, (b) CH 3 COO -, (c) H 2 PO 4 - HI (aq) H + (aq) + I - (aq)Brønsted acid CH 3 COO - (aq) + H + (aq) CH 3 COOH (aq)Brønsted base H 2 PO 4 - (aq) H + (aq) + HPO 4 2- (aq) H 2 PO 4 - (aq) + H + (aq) H 3 PO 4 (aq) Brønsted acid Brønsted base

A Lewis acid is a substance that can accept a pair of electrons A Lewis base is a substance that can donate a pair of electrons H+H+ H O H + OH - acidbase N H H H H+H+ + acidbase N H H H H + DEFINITION OF ACID AND BASE BY LEWIS

Examples of Lewis Acids and Bases reactions: N H H H acidbase F B F F + F F N H H H b) Ag + (aq) + 2NH 3 (aq)Ag(NH 3 ) 2+ (aq) acid base c) Cd + (aq) + 4I - (aq)CdI 2- 4 (aq) acid base d) Ni (s) + 4CO (g)Ni(CO) 4 (g) acid base a)

Acids i) Strong acids: -Acids that completely ionized in solution. -Example: HCl (aq) → H + (aq) + Cl - (aq) ii) Weak acids -Acids that incompletely ionized in solution -Example: CH 3 COOH (aq) CH 3 COO - (aq) + H + (aq) TYPES OF ACIDS-BASES

Monoprotic acid: - each unit of the acid yields one hydrogen ion upon ionization HCl H + + Cl - HNO 3 H + + NO 3 - CH 3 COOH H + + CH 3 COO - Strong electrolyte, strong acid Weak electrolyte, weak acid Diprotic acid: - each unit of the acid gives up two H + ions, in two separate steps H 2 SO 4 H + + HSO 4 - HSO 4 - H + + SO 4 2- Strong electrolyte, strong acid Weak electrolyte, weak acid

Triprotic acids: - yield three H + ions H 3 PO 4 H + + H 2 PO 4 - H 2 PO 4 - H + + HPO 4 2- HPO 4 2- H + + PO 4 3- Weak electrolyte, weak acid

Bases i) Strong bases: -Bases that completely ionized in solution. -Example: NaOH (s) → Na + (aq) + OH - (aq) ii) Weak bases -bases that incompletely ionized in solution -Example: NH 3 (aq) + H 2 O (l) NH 4 + (aq) + OH - (aq)

Acids and bases as electrolytes Strong acids such as HCl and HNO 3 are strong electrolytes, while weak acid such as acetic acid (CH 3 COOH) is a weak electrolyte. HCl (aq) + H 2 O (l) H 3 O + (aq) + Cl - (aq) HNO 3 (aq) + H 2 O (l) H 3 O + (aq) + NO 3 - (aq) HClO 4 (aq) + H 2 O (l) H 3 O + (aq) + ClO 4 - (aq) H 2 SO 4 (aq) + H 2 O (l) H 3 O + (aq) + HSO 4 - (aq)

Strong Acids are strong electrolytes HCl (aq) + H 2 O (l) H 3 O + (aq) + Cl - (aq) HNO 3 (aq) + H 2 O (l) H 3 O + (aq) + NO 3 - (aq) HClO 4 (aq) + H 2 O (l) H 3 O + (aq) + ClO 4 - (aq) H 2 SO 4 (aq) + H 2 O (l) H 3 O + (aq) + HSO 4 - (aq) Acids and bases as electrolytes HF (aq) + H 2 O (l) H 3 O + (aq) + F - (aq) Weak Acids are weak electrolytes HNO 2 (aq) + H 2 O (l) H 3 O + (aq) + NO 2 - (aq) HSO 4 - (aq) + H 2 O (l) H 3 O + (aq) + SO 4 2- (aq) H 2 O (l) + H 2 O (l) H 3 O + (aq) + OH - (aq)

Strong Bases are strong electrolytes NaOH (s) Na + (aq) + OH - (aq) H2OH2O KOH (s) K + (aq) + OH - (aq) H2OH2O Ba(OH) 2 (s) Ba 2+ (aq) + 2OH - (aq) H2OH2O F - (aq) + H 2 O (l) OH - (aq) + HF (aq) Weak Bases are weak electrolytes NO 2 - (aq) + H 2 O (l) OH - (aq) + HNO 2 (aq)

Conjugate acid-base pairs: The conjugate base of a strong acid has no measurable strength. H 3 O + is the strongest acid that can exist in aqueous solution. The OH - ion is the strongest base that can exist in aqueous solution.

H 2 O (l) H + (aq) + OH - (aq) autoionization of water Can act either as a acid or as a base. Water functions as a base with acids such as HCl and CH 3 COOH and function as acid in reaction with bases. Water is a very weak electrolyte and undergo ionization to a small extent: ACID-BASE PROPERTIES OF WATER

H 2 O (l) H + (aq) + OH - (aq) The Ion Product of Water K c = [H + ][OH - ] [H 2 O] [H 2 O] = constant K c = equilibrium constant K c [H 2 O] = K w = [H + ][OH - ] The ion-product constant (K w ) is the product of the molar concentrations of H + and OH - ions at a particular temperature. At 25 0 C K w = [H + ][OH - ] = 1.0 x [H + ] = [OH - ] [H + ] > [OH - ] [H + ] < [OH - ] Solution Is neutral acidic basic

What is the concentration of OH - ions in a HCl solution whose hydrogen ion concentration is 1.3 M? K w = [H + ][OH - ] = 1.0 x [H + ] = 1.3 M [OH - ] = KwKw [H + ] 1 x = = 7.7 x M

pH = - log [H + ] [H + ] = [OH - ] [H + ] > [OH - ] [H + ] < [OH - ] Solution Is neutral acidic basic [H + ] = 1 x [H + ] > 1 x [H + ] < 1 x pH = 7 pH < 7 pH > 7 At 25 0 C pH[H + ] pH-A MEASURE OF ACIDITY pH – the negative logarithm of the hydrogen in concentration (in mol/L)

pOH = -log [OH - ] [H + ][OH - ] = K w = 1.0 x log [H + ] – log [OH - ] = pH + pOH = Other important relationships pH Meter

1) The pH of rainwater collected in a certain region of the northeastern United States on a particular day was What is the H + ion concentration of the rainwater? pH = - log [H + ] [H + ] = 10 -pH = = 1.5 x M 2) The OH - ion concentration of a blood sample is 2.5 x M. What is the pH of the blood? pH + pOH = pOH = -log [OH - ]= -log (2.5 x )= 6.60 pH = – pOH = – 6.60 = 7.40

CALCULATION OF pH FOR SOLUTION CONTAINING A STRONG ACID AND A SOLUTION OF A STRONG BASE

1) What is the pH of a 2 x M HNO 3 solution? HNO 3 is a strong acid – 100% dissociation. HNO 3 (aq) + H 2 O (l) H 3 O + (aq) + NO 3 - (aq) pH = -log [H + ] = -log [H 3 O + ] = -log(0.002) = 2.7 Start End M 0.0 M 2) What is the pH of a 1.8 x M Ba(OH) 2 solution? Ba(OH) 2 is a strong base – 100% dissociation. Ba(OH) 2 (s) Ba 2+ (aq) + 2OH - (aq) Start End M M0.0 M pH = – pOH = log(0.036) = 12.6

HA (aq) + H 2 O (l) H 3 O + (aq) + A - (aq) Weak Acids (HA) and Acid Ionization Constants HA (aq) H + (aq) + A - (aq) K a = [H + ][A - ] [HA] K a is the acid ionization constant KaKa weak acid strength

1) What is the pH of a 0.5 M HF solution (at 25 0 C)? HF (aq) H + (aq) + F - (aq) K a = [H + ][F - ] [HF] = 7.1 x HF (aq) H + (aq) + F - (aq) Initial (M) Change (M) Equilibrium (M) x-x+x+x x x+x xx K a = x2x x = 7.1 x Ka  Ka  x2x = 7.1 x – x  0.50 K a << 1 x 2 = 3.55 x x = M [H + ] = [F - ] = M pH = -log [H + ] = 1.72 [HF] = 0.50 – x = 0.48 M

When can I use the approximation? 0.50 – x  0.50 K a << 1 When x is less than 5% of the value from which it is subtracted. x = M 0.50 M x 100% = 3.8% Less than 5% Approximation ok. 1) What is the pH of a 0.05 M HF solution (at 25 0 C)? Ka  Ka  x2x = 7.1 x x = M M 0.05 M x 100% = 12% More than 5% Approximation not ok. Must solve for x exactly using quadratic equation or method of successive approximations.

Solving weak acid ionization problems: 1.Identify the major species that can affect the pH. In most cases, you can ignore the autoionization of water. Ignore [OH - ] because it is determined by [H + ]. 2.Use ICE to express the equilibrium concentrations in terms of single unknown x. 3.Write K a in terms of equilibrium concentrations. Solve for x by the approximation method. If approximation is not valid, solve for x exactly. 4.Calculate concentrations of all species and/or pH of the solution.

1) What is the pH of a M monoprotic acid whose K a is 5.7 x ? HA (aq) H + (aq) + A - (aq) Initial (M) Change (M) Equilibrium (M) x-x+x+x x x+x xx K a = x2x x = 5.7 x Ka  Ka  x2x = 5.7 x – x  K a << 1 x 2 = 6.95 x x = M M M x 100% = 6.8% More than 5% Approximation not ok.

K a = x2x x = 5.7 x x x – 6.95 x = 0 ax 2 + bx + c =0 -b ± b 2 – 4ac  2a2a x = x = x = HA (aq) H + (aq) + A - (aq) Initial (M) Change (M) Equilibrium (M) x-x+x+x x x+x xx [H + ] = x = M pH = -log[H + ] = 2.09

NH 3 (aq) + H 2 O (l) NH 4 + (aq) + OH - (aq) Weak Bases and Base Ionization Constants K b = [NH 4 + ][OH - ] [NH 3 ] K b is the base ionization constant KbKb weak base strength Solve weak base problems like weak acids except solve for [OH-] instead of [H + ].

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 1) What mass of KI is required to make 500. mL of a 2.80 M KI solution? volume of KI solutionmoles KIgrams KI M KI 500. mL= 232 g KI 166 g KI 1 mol KI x 2.80 mol KI 1 L soln x 1 L 1000 mL x CONCENTRATION OF SOLUTION

Preparing a Solution of Known Concentration

Dilution is the procedure for preparing a less concentrated solution from a more concentrated solution. Dilution Add Solvent Moles of solute before dilution (i) Moles of solute after dilution (f) = MiViMiVi MfVfMfVf = DILUTION OF SOLUTIONS

EXAMPLE: 1) How would you prepare 60.0 mL of M HNO 3 from a stock solution of 4.00 M HNO 3 ? M i V i = M f V f M i = 4.00 M M f = MV f = L V i = ? L V i = MfVfMfVf MiMi = M x L 4.00 M = L = 3.00 mL Dilute 3.00 mL of acid with water to a total volume of 60.0 mL.

Concentration Units The concentration of a solution is the amount of solute present in a given quantity of solvent or solution. Percent by Mass (%w/w) % by mass = x 100% mass of solute mass of solute + mass of solvent = x 100% mass of solute mass of solution

Percent by Volume (%v/v) % by volume = x 100% Volume of solute Volume of solution Mole Fraction (X) X A = moles of A sum of moles of all components

M = moles of solute liters of solution Molarity (M) Molality (m) m = moles of solute mass of solvent (kg)

Quantitative analytical process based on measuring volumes. The most common form of VA is the titration, a process whereby a standard solution of known concentration is chemically reacted with a solution of unknown concentration in order to determine the concentration of the unknown. VOLUMETRIC ANALYSIS (VA)

In a titration a solution of accurately known concentration (standard solution) is added gradually added 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 TITRATIONS Titrations can be used in the analysis of acid-base reactions H 2 SO 4 + 2NaOH 2H 2 O + Na 2 SO 4

Slowly add base to unknown acid UNTIL the indicator changes color PROCEDURE FOR THE TITRATION

EXAMPLE: 1) What volume of a M NaOH solution is required to titrate mL of a 4.50 M H 2 SO 4 solution? WRITE THE CHEMICAL EQUATION! volume acidmoles redmoles basevolume base H 2 SO 4 + 2NaOH 2H 2 O + Na 2 SO mol H 2 SO mL soln x 2 mol NaOH 1 mol H 2 SO 4 x 1000 ml soln mol NaOH x mL = 158 mL M acid rxn coef. M base

ACID-BASE TITRATIONS Strong Acid-Strong Base Titrations NaOH (aq) + HCl (aq) H 2 O (l) + NaCl (aq) OH - (aq) + H + (aq) H 2 O (l) pH PROFILE OF THE TITRATION (TITRATION CURVE)

Before addition of NaOH - pH = 1.00 When the NaOH added - pH increase slowly at first Near the equivalence point (the point which equimolar amounts of acid and base have reacted) - the curve rises almost vertically Beyond the equivalence point - pH increases slowly pH PROFILE OF THE TITRATION (TITRATION CURVE)

CALCULATION OF Ph AT EVERY STAGE OF TITRATION 1)After addition of 10.0 mL of M NaOH to 25.0 mL of M HCl Total volume = 35.0 mL Moles of NaOH in 10.0 mL = 10.0 mL x mol NaOH x 1L 1L NaOH 1000 mL = 1.00 x mol Moles of HCl in 25.0 mL = 25.0 mL x mol HCl x 1L 1 L HCl 1000 mL = 2.50 x mol

Amount of HCl left after partial neutralization = (2.50 x )-(1.00 x ) = 1.50 x mol Concentration of H + ions in 35.0 mL 1.50 x mol HCl x 1000 mL = M HCl 35.0 mL 1L [H + ] = M, pH = -log = ) After addition of 25.0 mL of M NaOH to 25.0 mL 0f M HCl [H + ] = [OH - ] = 1.00 x 10-7 pH = 7.00

3) After addition of 35.0 mL of M NaOH to 25.0 mL of mL of HCl Total volume = 60.0 mL Moles of NaOH added = 35.0 mL x mol NaOH x 1L 1 L NaOH 000 mL = 3.50 x mol Moles of HCl in 25.0 mL solution = 2.50 x mol After complete neutralization of HCl, no of moles of NaOH left = (3.50 x )-(2.50x10 -3 ) = 1.00 x mol Concentration of NaOH in 60.0 mL solution = 1.00 x mol NaOH x 1000 mL 60.0 mL 1L = M NaOH [OH - ] = M pOH = -log = 1.78 pH = – 1.78 = 12.22

Exactly 100 mL of 0.10 M HNO 2 are titrated with a 0.10 M NaOH solution. What is the pH at the equivalence point ? HNO 2 (aq) + OH - (aq) NO 2 - (aq) + H 2 O (l) start (moles) end (moles) NO 2 - (aq) + H 2 O (l) OH - (aq) + HNO 2 (aq) Initial (M) Change (M) Equilibrium (M) x-x+x+x x x+x xx [NO 2 - ] = = 0.05 M Final volume = 200 mL K b = [OH - ][HNO 2 ] [NO 2 - ] = x2x x = 2.2 x – x  0.05x  1.05 x = [OH - ] pOH = 5.98 pH = 14 – pOH = 8.02

Acid-Base Indicators HIn (aq) H + (aq) + In - (aq)  10 [HIn] [In - ] Color of acid (HIn) predominates  10 [HIn] [In - ] Color of conjugate base (In - ) predominates