Acids and Bases RNA uses amino-acids to build proteins/enzymes

Acids and Bases RNA uses amino-acids to build proteins/enzymes
Digestive Acids help to break down food into reusable molecular fragments Acids and Bases It is the acids in citrus fruits that give them the sour taste and allows the fruit to stay in a state of preservation till germination

Properties of Acids sour taste
react with active metals (Al, Zn, Fe), but not Cu, Ag, or Au 2 Al(s) + 6 HCl(aq)  2 AlCl3(aq) + 3 H2(g) Corrosive react with carbonates, producing CO2 marble, baking soda, chalk, limestone CaCO3(s) + 2 HCl(aq)  CaCl2(aq) + CO2(g) + H2O(l) change color of vegetable dyes blue litmus turns red react with bases to form ionic salts HCl(aq) + NaOH(aq)NaCl(aq) + H2O(l)

Common Acids

Structure of Acids binary acids have acid hydrogens attached to a nonmetal atom HCl, HF

Structure of Acids oxy acids have acid hydrogens attached to an oxygen atom H2SO4, HNO3

Structure of Acids carboxylic acids have COOH group HC2H3O2, H3C6H5O7
only the first H in the formula is acidic the H is on the COOH

Properties of Bases also known as alkalis taste bitter
alkaloids = plant product that is alkaline often poisonous solutions feel slippery change color of vegetable dyes different color than acid red litmus turns blue react with acids to form ionic salts Neutralization HCl(aq) + NaOH(aq)NaCl(aq) + H2O(l)

Common Bases

Structure of Bases most ionic bases contain OH- ions
NaOH, Ca(OH)2 some contain CO32- ions CaCO3 NaHCO3 molecular bases contain structures that react with H+ mostly amine groups Amino acids have a base at one end and an acid at the other, neighboring amino acids can neutralize to form a polypeptide

Indicators chemicals which change color depending on the acidity/basicity many vegetable dyes are indicators anthocyanins litmus from Spanish moss red in acid, blue in base phenolphthalein found in laxatives red in base, colorless in acid Anthocyanins give these pansies their dark purple pigmentation and are the pigment in red cabbage that is so sensitive to acidity

acids and bases: Arrhenius Theory
bases dissociate in water to produce OH- ions and cations ionic substances dissociate in water NaOH(aq) → Na+(aq) + OH–(aq) acids ionize in water to produce H+ ions and anions HCl(aq) → H+(aq) + Cl–(aq) HC2H3O2(aq) H+(aq) + C2H3O2–(aq)

Arrhenius Theory HCl ionizes in water producing H+ and Cl– ions
NaOH dissociates in water producing Na+ and OH– ions

Hydronium Ion the H+ ions produced by the acid are so reactive they cannot exist in water H+ ions are protons instead, they react with a water molecule(s) to produce complex ions, mainly hydronium ion, H3O+ H+ + H2O  H3O+ ≅ H+(aq) there are also minor amounts of H+ with multiple water molecules, H(H2O)n+

Arrhenius Acid-Base Reactions
the H+ from the acid combines with the OH- from the base to make a molecule of H2O it is often helpful to think of H2O as H-OH the cation from the base combines with the anion from the acid to make a salt acid + base → salt + water HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) H+(aq)+Cl-(aq)+Na+(aq)+OH-(aq)Na+(aq)+Cl-(aq)+H2O(l) H+(aq) + OH-(aq) H2O(l) All acid base reactions have this same net ionic equation in the Arrhenius idea of the acid and base

Limitations of the Arrhenius Theory
does not explain why molecular substances, like NH3, dissolve in water to form basic solutions – even though they do not contain OH– ions does not explain how some ionic compounds, like Na2CO3 or Na2O, dissolve in water to form basic solutions – even though they do not contain OH– ions does not explain why molecular substances, like CO2, dissolve in water to form acidic solutions – even though they do not contain H+ ions does not explain acid-base reactions that take place outside aqueous solution

Acids and bases: Brønsted-Lowry
in a Brønsted-Lowry Acid-Base reaction, an H+ is transferred does not have to take place in aqueous solution broader definition than Arrhenius An acid is H+ donor, base is H+ acceptor base structure must contain an atom with an unshared pair of electrons in an acid-base reaction, the acid molecule gives an H+ to the base molecule H–A + :B  :A– + H–B+

Brønsted-Lowry Acids HCl(aq) + H2O(l) → Cl–(aq) + H3O+(aq)
Brønsted-Lowry acids are H+ donors any material that has H can potentially be a Brønsted-Lowry acid because of the molecular structure, often one H in the molecule is easier to transfer than others HCl(aq) is acidic because HCl transfers an H+ to H2O, forming H3O+ ions water acts as base, accepting H+ HCl(aq) + H2O(l) → Cl–(aq) + H3O+(aq) Acid base

Brønsted-Lowry Bases NH3(aq) + H2O(l)  NH4+(aq) + OH–(aq)
Brønsted-Lowry bases are H+ acceptors any material that has atoms with lone pairs can potentially be a Brønsted-Lowry base because of the molecular structure, often one atom in the molecule is more willing to accept H+ transfer than others NH3(aq) is basic because NH3 accepts an H+ from H2O, forming OH–(aq) water acts as acid, donating H+ NH3(aq) + H2O(l)  NH4+(aq) + OH–(aq) base acid Tro, Chemistry: A Molecular Approach

Amphoteric Substances
amphoteric substances can act as either an acid or a base have both transferable H and atom with lone pair Example water acts as base, accepting H+ from HCl HCl(aq) + H2O(l) → Cl–(aq) + H3O+(aq) water acts as acid, donating H+ to NH3 NH3(aq) + H2O(l) → NH4+(aq) + OH–(aq)

Brønsted-Lowry: Acid-Base Reactions
one of the advantages of Brønsted-Lowry theory is that it allows reactions to be reversible H–A + :B :A– + H–B+ the original base has an extra H+ after the reaction – so it will act as an acid in the reverse process and the original acid has a lone pair of electrons after the reaction – so it will act as a base in the reverse process :A– + H–B H–A + :B

Conjugate Pairs In a Brønsted-Lowry Acid-Base reaction, the original base becomes an acid in the reverse reaction, and the original acid becomes a base in the reverse process each reactant and the product it becomes is called a conjugate pair the original base becomes the conjugate acid; and the original acid becomes the conjugate base NH3(aq) + H2O(l) NH4+(aq) OH–(aq) Base Acid Conjugate Acid Conjugate Base

Brønsted-Lowry: Acid-Base Reactions
H–A :B :A– H–B+ acid base conjugate conjugate base acid HCHO H2O CHO2– H3O+ acid base conjugate conjugate base acid H2O NH HO– NH4+ acid base conjugate conjugate base acid

Conjugate Pairs In the reaction H2O + NH3 HO– + NH4+
H2O and HO– constitute an Acid/Conjugate Base pair NH3 and NH4+ constitute a Base/Conjugate Acid pair

acid base conjugate conjugate base acid
Identify the Brønsted-Lowry Acids and Bases and Their Conjugates in the Reaction H2SO H2O HSO4– + H3O+ When the H2SO4 becomes HSO4-, it lost an H+ so H2SO4 must be the acid and HSO4- its conjugate base When the H2O becomes H3O+, it accepted an H+ so H2O must be the base and H3O+ its conjugate acid H2SO H2O HSO4– + H3O+ acid base conjugate conjugate base acid

base acid conjugate conjugate
Identify the Brønsted-Lowry Acids and Bases and Their Conjugates in the Reaction HCO3– H2O H2CO3 + HO– When the HCO3 becomes H2CO3, it accepted an H+ so HCO3- must be the base and H2CO3 its conjugate acid When the H2O becomes OH-, it donated an H+ so H2O must be the acid and OH- its conjugate base HCO3– H2O H2CO3 + HO– base acid conjugate conjugate acid base

Practice – Write the formula for the conjugate acid of the following
H2O NH3 CO32− H2PO4−

Practice – Write the formula for the conjugate acid of the following
H2O H3O+ NH3 NH4+ CO32− HCO3− H2PO41− H3PO4

Practice – Write the formula for the conjugate base of the following
H2O NH3 CO32− H2PO4−

Practice – Write the formula for the conjugate base of the following
H2O HO− NH3 NH2− CO32− since CO32− does not have an H, it cannot be an acid H2PO41− HPO42−

Arrow Conventions chemists commonly use two kinds of arrows in reactions to indicate the degree of completion of the reactions a single arrow indicates all the reactant molecules are converted to product molecules at the end a double arrow indicates the reaction stops when only some of the reactant molecules have been converted into products

Strong or Weak a strong acid is a strong electrolyte
practically all the acid molecules ionize, → a strong base is a strong electrolyte practically all the base molecules form OH– ions, either through dissociation or reaction with water, → a weak acid is a weak electrolyte only a small percentage of the molecules ionize, a weak base is a weak electrolyte only a small percentage of the base molecules form OH– ions, either through dissociation or reaction with water,

Strong Acids HCl(aq)  H+(aq) + Cl-(aq)
The stronger the acid, the more willing it is to donate H use water as the standard base strong acids donate practically all their H’s 100% ionized in water strong electrolyte [H3O+] = [strong acid] HCl(aq)  H+(aq) + Cl-(aq) HCl(aq) + H2O(l)  H3O+(aq)+ Cl-(aq)

Weak Acids HF(aq) H+(aq) + F-(aq) HF(aq) + H2O(l) H3O+(aq) + F-(aq)
weak acids donate a small fraction of their H’s most of the weak acid molecules do not donate H to water much less than 1% ionized in water [H3O+] << [weak acid] HF(aq) H+(aq) + F-(aq) HF(aq) + H2O(l) H3O+(aq) + F-(aq)

Polyprotic Acids often acid molecules have more than one ionizable H – these are called polyprotic acids the ionizable H’s may have different acid strengths or be equal 1 H = monoprotic, 2 H = diprotic, 3 H = triprotic HCl = monoprotic, H2SO4 = diprotic, H3PO4 = triprotic

Polyprotic Acids polyprotic acids ionize in steps
each ionizable H removed sequentially removing of the first H automatically makes removal of the second H harder H2SO4 is a stronger acid than HSO4-

Increasing Basicity Increasing Acidity

Strengths : Acids and Bases
commonly, Acid or Base strength is measured by determining the equilibrium constant of a substance’s reaction with water HA + H2O A-1 + H3O+1 B: + H2O HB+1 + OH-1 the farther the equilibrium position lies to the products, the stronger the acid or base the position of equilibrium depends on the strength of attraction between the base form and the H+ stronger attraction means stronger base or weaker acid

General Trends: Acidity
the stronger an acid is at donating H, the weaker the conjugate base is at accepting H higher oxidation number = stronger oxyacid H2SO4 > H2SO3; HNO3 > HNO2 cation stronger acid than neutral molecule; neutral stronger acid than anion H3O+1 > H2O > OH-1; NH4+1 > NH3 > NH2-1 base trend opposite

Acid Ionization Constant, Ka
acid strength measured by the size of the equilibrium constant when react with H2O HA + H2O A-1 + H3O+1 the equilibrium constant is called the acid ionization constant, Ka larger Ka = stronger acid

Autoionization of Water
Water is actually an extremely weak electrolyte therefore there must be a few ions present about 1 out of every 10 million water molecules form ions through a process called autoionization H2O H+ + OH– H2O + H2O H3O+ + OH– all aqueous solutions contain both H3O+ and OH– the concentration of H3O+ and OH– are equal in water [H3O+] = [OH–] = 25°C

Ion Product of Water the product of the H3O+ and OH– concentrations is always the same number the number is called the ion product of water and has the symbol Kw [H3O+] x [OH–] = Kw = 1 x 25°C if you measure one of the concentrations, you can calculate the other as [H3O+] increases the [OH–] must decrease so the product stays constant inversely proportional

Acidic and Basic Solutions
all aqueous solutions contain both H3O+ and OH– ions neutral solutions have equal [H3O+] and [OH–] [H3O+] = [OH–] = 1 x 10-7 acidic solutions have a larger [H3O+] than [OH–] [H3O+] > 1 x 10-7; [OH–] < 1 x 10-7 basic solutions have a larger [OH–] than [H3O+] [H3O+] < 1 x 10-7; [OH–] > 1 x 10-7

The units are correct. The fact that the
Calculate the [OH-] at 25°C when the [H3O+] = 1.5 x 10-9 M, and determine if the solution is acidic, basic, or neutral [H3O+] = 1.5 x 10-9 M [OH-] if [OH-] > [H3O+] basic if [H3O+]>[OH-] acid Given : Find: Concept Plan: Relationships: [H3O+] [OH-] Solution: Check: The units are correct. The fact that the [H3O+] < [OH-] means the solution is basic

Complete the Table [H+] vs. [OH-]

Complete the Table [H+] vs. [OH-]
Acid Base [H+] OH- H+ [OH-] Even though it may look like it, neither H+ nor OH- will ever be 0

Since in water [H3O+] = 10-7M
pH the acidity/basicity of a solution is often expressed as pH pH = -log[H3O+], [H3O+] = 10-pH Since in water [H3O+] = 10-7M pHwater = -log[10-7] = 7 pH < 7 is acidic; pH > 7 is basic, pH = 7 is neutral

pH the lower the pH, the more acidic the solution
the higher the pH, the more basic the solution 1 pH unit corresponds to a factor of 10 difference in acidity normal range 0 to 14 pH 0 is [H+] = 1 M, pH 14 is [OH–] = 1 M pH can be negative (very acidic) or larger than 14 (very alkaline)

Sig. Figs. and Logs when you take the log of a number written in scientific notation, the digit(s) before the decimal point come from the exponent on 10, and the digits after the decimal point come from the decimal part of the number log(2.0 x 106) = log(106) + log(2.0) = … = since the part of the scientific notation number that determines the significant figures is the decimal part, the sig figs are the digits after the decimal point in the log log(2.0 x 106) = 6.30

Tro, Chemistry: A Molecular Approach

Calculate the pH at 25°C when the [OH-] = 1
Calculate the pH at 25°C when the [OH-] = 1.3 x 10-2 M, and determine if the solution is acidic, basic, or neutral Given: Find: [OH-] = 1.3 x 10-2 M pH Concept Plan: Relationships: [H3O+] [OH-] pH Solution: Check: pH is unitless. The fact that the pH > 7 means the solution is basic

pOH another way of expressing the acidity/basicity of a solution is pOH pOH = -log[OH-], [OH-] = 10-pOH pOHwater = -log[10-7] = 7 need to know the [OH-] concentration to find pOH pOH < 7 is basic; pOH > 7 is acidic, pOH = 7 is neutral Tro, Chemistry: A Molecular Approach

pH and pOH Complete the Table

Relationship between pH and pOH
the sum of the pH and pOH of a solution = 14.00 at 25°C can use pOH to find pH of a solution

pK a way of expressing the strength of an acid or base is pK
pKa = -log(Ka), Ka = 10-pKa pKb = -log(Kb), Kb = 10-pKb the stronger the acid, the smaller the pKa larger Ka = smaller pKa because it is the –log

Finding the pH of a Strong Acid
there are two sources of H3O+ in an aqueous solution of a strong acid – the acid and the water for the strong acid, the contribution of the water to the total [H3O+] is negligible for a monoprotic strong acid [H3O+] = [HA] 0.10 M HCl has [H3O+] = 0.10 M and pH = 1.00

Finding the pH of a Weak Acid
there are also two sources of H3O+ in and aqueous solution of a weak acid – the acid and the water however, finding the [H3O+] is complicated by the fact that the acid only undergoes partial ionization calculating the [H3O+] requires solving an equilibrium problem for the reaction that defines the acidity of the acid Tro, Chemistry: A Molecular Approach

Find the pH of 0.200 M HNO2(aq) solution @ 25°C
Write the reaction for the acid with water Construct an ICE table for the reaction Enter the initial concentrations – assuming the [H3O+] from water is ≈ 0 HNO2 + H2O NO2- + H3O+ [HNO2] [NO2-] [H3O+] initial change equilibrium [HNO2] [NO2-] [H3O+] initial 0.200 ≈ 0 change equilibrium since no products initially, Qc = 0, and the reaction is proceeding forward

represent the change in the concentrations in terms of x sum the columns to find the equilibrium concentrations in terms of x substitute into the equilibrium constant expression [HNO2] [NO2-] [H3O+] initial 0.200 change equilibrium -x +x +x x x x

Ka for HNO2 = 4.6 x 10-4 determine the value of Ka from Table 15.5 since Ka is very small, approximate the [HNO2]eq = [HNO2]init and solve for x [HNO2] [NO2-] [H3O+] initial 0.200 ≈ 0 change -x +x equilibrium x x

Ka for HNO2 = 4.6 x 10-4 check if the approximation is valid by seeing if x < 5% of [HNO2]init [HNO2] [NO2-] [H3O+] initial 0.200 ≈ 0 change -x +x equilibrium x x = 9.6 x 10-3 the approximation is valid

Ka for HNO2 = 4.6 x 10-4 substitute x into the equilibrium concentration definitions and solve [HNO2] [NO2-] [H3O+] initial 0.200 ≈ 0 change -x +x equilibrium 0.190 0.0096 [HNO2] [NO2-] [H3O+] initial 0.200 ≈ 0 change -x +x equilibrium 0.200-x x x = 9.6 x 10-3

Ka for HNO2 = 4.6 x 10-4 substitute [H3O+] into the formula for pH and solve [HNO2] [NO2-] [H3O+] initial 0.200 ≈ 0 change -x +x equilibrium 0.190 0.0096

Ka for HNO2 = 4.6 x 10-4 check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated Ka to the given Ka [HNO2] [NO2-] [H3O+] initial 0.200 ≈ 0 change -x +x equilibrium 0.190 0.0096 though not exact, the answer is reasonably close

What is the pH of a 0.012 M solution of nicotinic acid, HC6H4NO2 given Ka = 1.4 x 10-5 @ 25°C?
Tro, Chemistry: A Molecular Approach

HC6H4NO2 + H2O C6H4NO2- + H3O+ Write the reaction for the acid with water Construct an ICE table for the reaction Enter the initial concentrations – assuming the [H3O+] from water is ≈ 0 [HA] [A-] [H3O+] initial 0.012 ≈ 0 change equilibrium

HC6H4NO2 + H2O C6H4NO2- + H3O+ [HA] [A-] [H3O+] initial 0.012 change equilibrium represent the change in the concentrations in terms of x sum the columns to find the equilibrium concentrations in terms of x substitute into the equilibrium constant expression -x +x +x x x x

HC6H4NO2 + H2O C6H4NO2- + H3O+ determine the value of Ka since Ka is very small, approximate the [HA]eq = [HA]init and solve for x [HA] [A2-] [H3O+] initial 0.012 ≈ 0 change -x +x equilibrium x x

Ka for HC6H4NO2 = 1.4 x 10-5 check if the approximation is valid by seeing if x < 5% of [HC6H4NO2]init [HA] [A2-] [H3O+] initial 0.012 ≈ 0 change -x +x equilibrium x x = 4.1 x 10-4 the approximation is valid

initial 0.012 ≈ 0 change -x +x equilibrium 0.012-x x substitute x into the equilibrium concentration definitions and solve x = 4.1 x 10-4

substitute [H3O+] into the formula for pH and solve [HA] [A2-] [H3O+] initial 0.012 ≈ 0 change -x +x equilibrium

check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated Ka to the given Ka [HA] [A2-] [H3O+] initial 0.012 ≈ 0 change -x +x equilibrium the values match

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