Chemistry: The Molecular Science

Name: Chemistry: The Molecular Science
Uploaded: 2017-07-04T10:24:19+00:00
Duration: PTM45S3
Channel: Eddie Stubblefield
Description: Chemistry: The Molecular Science

Chemistry: The Molecular Science
Moore, Stanitski and Jurs Chapter 16: Acids and Bases

Arrhenius Definition Arrhenius: any substance which ionizes in water to produce… protons (H+) is an acid. hydroxide ions (OH-) is a base. Better version of the Arrhenius definition: Acid: hydronium ions (H3O+) in water are acidic. Base: hydroxide ions (OH-) in water are basic.

Acid-Base Reactions A strong acid is an acid that ionizes completely in water (product favored). There are six strong acids: HCl - hydrochloric acid HBr – hydrobromic acid HI – hydroiodic acid H2SO4 - sulfuric acid HNO3 – nitric acid HClO4 – perchloric acid

Acid-Base Reactions Strong Acids and Bases
A strong base is a base that is present entirely as ions (product favored). The hydroxides of Group IA, IIA (except Be and Mg hydroxides) are strong bases. So, why is NH3 (aq) basic?

Brønsted-Lowry Concept
An alternative definition: acid = proton (H+) donor base = proton (H+) acceptor Works for non-aqueous solutions and explains why NH3 is basic: NH3(g) + H2O(l) NH4+(aq) + OH-(aq) indirectly produced Base: H+ acceptor Acid: H+ donor

Brønsted-Lowry Concept
Strong acids and bases almost completely ionize (product favored). 100% HNO3(aq) + H2O(l) H3O+(aq) + NO3-(aq) Weak acids and bases do not fully ionize (reactant favored). 3% HF(aq) + H2O(l) H3O+(aq) + F-(aq) Acid: H+ donor Base: H+ acceptor Acid: H+ donor Base: H+ acceptor Note: the products are a new acid and base pair. LeChatelier – a more dilute solution (more water), will ionize more.

Water’s Role as Acid or Base
Water acts as a base when an acid dissolves in water: HBr(aq) + H2O(l) H3O+(aq) + Br-(aq) acid base acid base But water acts as an acid for some bases: H2O(l) + NH3(aq) NH4+(aq) + OH-(aq) acid base acid base Water is amphiprotic - it can donate or accept a proton (act as acid or base).

Practice Complete the equations: (acid or base?, mass/charge balance?, single or double arrow?) HClO4 + H2O CN- + H2O H2S + H2O NO H2O Ca(OH)2 + H2O ClO H3O+ HCN + HO- HS- + H3O+ HNO2 + HO- Ca OH-

Conjugate Acid-Base Pairs
Molecules or ions related by the loss/gain of one H+. Conjugate Acid Conjugate Base H3O+ H2O CH3COOH CH3COO- NH4+ NH3 H2SO4 HSO4- HSO4- SO42- HCl Cl- donate H+ accept H+ NH4+ and NH2- are not conjugate (conversion requires 2 H+)

Conjugate Acid-Base Pairs
Identify the base conjugate to HC2H3O2(aq) and the acid conjugate to HCO3-(aq) HC2H3O H2O H3O+ + C2H3O2 - Conjugate Base Acid: H+ donor HCO H2O OH- + H2CO3 Base: H+ acceptor Conjugate Acid 160 of 200 Show that HCO3- is amphoteric.

Practice: Conjugate Acid-Base Pairs
Write the equation for the acid or base in water and identify the acid-base pairs. HBr HSO3- (as an acid) HSO3- (as a base) PH4+

Relative Strength of Acids & Bases
Strong acids are better H+ donors than weak acids. Strong bases are better H+ acceptors than weak bases. Strong acids have weak conjugate bases. Weak acids have strong conjugate bases. Strong acid + H2O H3O+ + weak conjugate base Fully ionized, reverse reaction essentially does not occur. The conjugate base is weak. Weak acid + H2O H3O+ + strong conjugate base Weakly ionized, reverse reaction readily occurs. The conjugate base is strong.

Conjugate acid Conjugate base H2SO4 HSO4- HBr Br- HCl Cl- HNO3 NO3- H3O+ H2O H2SO3 HSO3- HSO4- SO42- H3PO4 H2PO4- HF F- CH3COOH CH3COO- H2S HS- H2PO4- HPO42- NH4+ NH3 HCO3- CO32- H2O OH- OH- O2- H2 H- CH4 CH3- Acid strength increasing Base strength increasing stong acids strong base extremely weak acids weak bases

Problem Is the following aqueous reaction product or reactant favored? Acid strength increasing Base strength increasing Conj acid. Conj. base H2SO4 HSO4- HBr Br- HCl Cl- HF F- NH4+ NH3 OH- O2- H2 H- CH4 CH3- NH4+ + F NH3 + HF HF is a stronger acid than NH4+. (NH3 is a stronger base than F-) HF has greater tendency to ionize than NH4+. (NH3 more readily accepts H+ than F-) Reactant Favored

Carboxylic Acids C3H7COOH CH3COOH

Amines (bases) proton acceptors CH3NH2 (CH3)2NH (CH3)3N

Autoionization of Water
Pure water conducts a very small electrical current. Autoionization occurs: H2O(l) + H2O(l) H3O+(aq) + OH-(aq) Base Acid Acid Base Kw = [ H3O+ ] [ OH- ] Kw = 1.0 x (at 25°C) Kw = ionization constant for water Heavily reactant favored.

Ionization Constant for Water
Kw is a special equilibrium constant for the autoionization of water. It is T-dependent. T = 25°C (77°F) is usually used as the standard T. T (°C) Kw x 10-14 x 10-14 x 10-14 x 10-14 x 10-14 x 10-14 Kw increases with temperature. Is it endo or exothermic?

Hydronium and hydroxide ions are produced in equal numbers in pure water, 2H2O(l) H3O+(aq) + OH-(aq) Kw = [ H3O+ ] [ OH- ] Thus, the concentrations of H3O+ and OH- in pure water are both 1.0 x 10-7 M. 2

H3O+ and OH- are present in all aqueous solutions. Neutral solution. Pure water 25°C): [H3O+] = 10-7 M = [OH-] Acidic solution If acid is added to water: [ H3O+ ] is increased, disturbing the equilibrium: 2 H2O H3O+ + OH- LeChatelier: equilibrium shifts to the left. [OH-] Equilibrium is reestablished: [ H3O+ ] > 10-7 M > [OH-] Product of ion concentrations is the same [ H3O+ ][OH-] = Kw = 1.0 x 10-14

Basic solution. If base is added to water. LeChatelier: equilibrium shifts to the left, [H3O+] New equilibrium: [H3O+] < 10-7 M < [OH-] [ H3O+ ][OH-] = Kw = 1.0 x at 25°C 2 H2O H3O+ + OH- Example Calculate the hydronium and hydroxide ion concentrations at 25°C in a 6.0 M aqueous sodium hydroxide solution. Kw = [H3O+][OH-] = 1.0 x 10-14 NaOH (aq) is strong (100% ionized) so [OH-] = 6.0 M. [H3O+](6.0) = 1.0 x 10-14 [H3O+] = 1.7 x M [ OH- ] = 6.0 M

The relative concentrations of H3O+ and OH- indicate acidic, neutral or basic character of a solution: In a neutral solution, the concentrations of H3O+ (aq) and OH-(aq) are equal. In an acidic solution, the concentration of H3O+ (aq) is greater than that of OH-(aq). In a basic solution, the concentration of OH-(aq) is greater than that of H3O+ (aq). 2

Acid or Base At 25 oC, we observe the following conditions.
In an acidic solution, [H3O+ ] > 1.0 x 10-7 M. In a neutral solution, [H3O+ ] = [OH-] = 1.0 x 10-7 M. In a basic solution, [H3O+ ] < 1.0 x 10-7 M. 2

The pH scale pH = −log10[H3O+]
Can describe acidity in terms of [H3O+], but a logarithmic scale is more convenient: pH = −log10[H3O+] At 25°C a neutral aqueous solution has: pH = −log10[1.0 x 10-7] = −(−7.00) = 7.00 acidic solutions: [H3O+]>1.0 x 10-7, pH < 7.00 basic solutions: [H3O+]<1.0 x 10-7, pH > 7.00

The pH of a Solution For a solution in which the hydronium-ion concentration is 1.0 x 10-3, the pH is: Note that the number of decimal places in the pH equals the number of significant figures in the hydronium-ion concentration. 2

Example – pH given [H3O+] ?
A sample of orange juice has a hydronium-ion concentration of 2.9 x 10-4 M. What is the pH? 2

Example - [H3O+] given pH?
The pH of human arterial blood is 7.40 (acidic or basic?). What is the hydronium-ion concentration? (Guesss….> or < 1.0x10-7M?) 10logx = x 2

The pOH scale pOH = −log10[OH-] A neutral solution (25°C) has:
pOH = −log10[1.0 x 10-7] = −(−7.00) = 7.00 Since Kw = [ H3O+ ][ OH- ] = 1.0 x 10-14 −log(KW)= −log[H3O+] + (−log[OH-]) = −log(1.0 x 10-14) pKw = pH + pOH = 14.00

we first calculate the pOH:
Example An ammonia solution has a hydroxide-ion concentration of 1.9 x 10-3 M. What is the pH of the solution? we first calculate the pOH: Then the pH is: 2

pH Calculations Given two aqueous solutions (25°C).
Solution A: [OH-] = 4.3 x 10-4 M, Solution B: [H3O+] = 7.5 x 10-9 M. Which has the higher pH? Which is more acidic? Solution A: pOH = −log[ OH-] = 3.37 pH + pOH = pKw = pH = Solution B: pH = −log[ H3O+] = 8.12 A has higher pH, B is more acidic

Practice - WSs [H+] Kw =1x10-14 = [H+] [OH-] [OH-] pH= -log [H+]
pOH =-log [OH-] [OH-] = 10-pOH pH = pH + pOH pOH

pH of Aqueous Solutions
Stomach Fluids Lemon Juice Vinegar Wine Tomatoes Black Coffee Milk Pure water Blood, seawater Sodium bicarbonate Borax solution Milk of Magnesia Detergents Aqueous ammonia 1 M NaOH [H3O+] [OH-] Example Battery Acid pH 1 2 3 4 5 6 7 8 9 10 11 12 13 14 Bleach pH of Aqueous Solutions

Measuring pH H3O+ concentrations can be measured with an:
Electronic pH meter: fast and accurate preferred method. Acid-base Indicator: substance that changes color within a narrow pH range may have multiple color change (e.g. bromthymol blue) one “color” may be colorless (e.g. phenolphthalein) cheap and convenient.

Ionization Constants of Acids and Bases
When an acid ionizes in water: HA(aq) + H2O(l) H3O+(aq) + A-(aq) The acid ionization constant is used to report the degree of ionization: [A-][H3O+] [HA] Ka = (Water omitted, as usual) Strong acids have large Ka values Weak acid have small Ka values

Base Ionization Constants
For a base in water: B(aq) + H2O(l) BH+(aq) + OH-(aq) The base ionization constant, Kb, is: Kb = [BH+][OH-] [B] If the base is an anion: A-(aq) + H2O(l) HA(aq) + OH-(aq) Kb = [HA][OH-] [A-]

Ionization Constants Larger Ka = stronger acid
Larger Kb = stronger base

Acid and Base Strength Small value of Ka indicates that the numerator is smaller than the denominator. Equilibrium favors reactants or only a small amount dissociates – a weak acid. Small value of Kb indicates that the numerator is smaller than the denominator. Equilibrium favors reactants or only a small amount dissociates – a weak base.

Example Write the ionization equation and ionization constant expression for the following acids and bases: a.) HF b.) HBrO c.)NH3 d.) CH3NH2

Ka Values for Polyprotic Acids
Some acids can donate more than one H+ Formula Name Acidic H’s H2S Hydrosulfuric Acid 2 H3PO4 Phosphoric Acid 3 H2CO3 Carbonic Acid 2 HOOC-COOH Oxalic acid 2 C3H5(COOH)3 Citric acid 3 Each H+ ionization has a different Ka. The 1st proton is easiest to remove The 2nd is harder, etc.

Ka Values for Polyprotic Acids
Phosphoric acid (H3PO4) is weak. It has three acidic protons: H3PO4(aq) + H2O(l) H3O+(aq) + H2PO4- (aq) Ka = 7.5 x 10-3 H+ is harder to remove Ka decreasing H2PO4-(aq) + H2O(l) H3O+(aq) + HPO42- (aq) Ka = 6.2 x 10-8 HPO42-(aq) + H2O(l) H3O+(aq) + PO43- (aq) Ka = 3.6 x 10-13

Practice – Relative Strength (Ka and Kb)
The following reactions all have Keq>1. NO HF HNO2 + F- CH3COO- + HF CH3COOH + F- HNO2 + CH3COO HNO CH3COOH Arrange the substances based on their relative base strength. F- CH3COOH HF HNO2- CH3COO- HNO2 3 4 2 1 1 – strongest base 2 – intermediate base 3 – weakest base 4 – not a B-L base

Molecular Structure and Acid Strength
What makes a strong acid? A weak H–A bond, so H+ can be easily removed! Consider the binary acids HF, HCl, HBr, HI. HX Bond Energy (kJ) Ka HF weak acid x 10-4 HCl x 107 HBr x 108 HI x 1010 strong acids smaller bond energy larger acid strength H-A, bond energy decreases down a group. Acid strength increases.

What makes a strong acid? A weak H–A bond, so H+ can be easily removed! Consider the binary acids SiH4, PH3, H2S, HCl. The higher electronegativity, across a period, makes the bond more polar and hence more ionic – a stronger acid.

Oxoacids hydrogen, oxygen and one other element H O-E higher the electronegativity on E, stronger the acid as this weakens the bond between the O and H HOCl HOBr HOI Ka: × × × 10-11 Electronegativity: Cl = 3.0, Br = 2.8, I = 2.5

The acid strength increases as the number of oxygen atoms bonded to E increases. Strong oxoacid has at least two oxygen atoms per acidic H atoms. HClO HClO HClO *HClO4 Ka: 3.5 × × ≈ ≈ 108

Problem Solving Using Ka and Kb
Example: Ka from pH Lactic acid is monoprotic. The pH of a M solution was 2.43 at 25°C. Determine Ka for this acid. Using the pH information: - log [H3O+] = 2.43 [H3O+] = = M These hydronium ions are produced by: HA(aq) + H2O(l) H3O+(aq) + A-(aq) and from the autoionization of water: H2O(l) + H2O(l) H3O+(aq) + OH-(aq)

HA(aq) + H2O(l) H3O+(aq) + A-(aq) [ ]initial x (water autoionization) change [ ]equil – Ka = [H3O+][A-] [HA] Ka = = 1.4 x 10-4 (0.0037)(0.0037) (0.100 – )

Example: pH from Ka Determine the pH of a M propanoic acid solution at 25°C. Ka = 1.4 x What % of acid is ionized? C2H5COOH + H2O(l) H3O+(aq) + C2H5COO-(aq) Ka = = 1.4 x 10-5 [H3O+][C2H5COO-] [C2H5COOH]

Determine the pH of M propanoic acid soln. at 25°C. Ka = 1.4 x What % of acid is ionized? HA(aq) + H2O(l) H3O+(aq) A-(aq) [ ]initial * 0 change -x x x [ ]equil – x x* x *Ignore the water contribution [H3O+][A-] [HA] Ka = 1.4 x 10-5 = 1.4 x 10-5= x2 (0.100 – x)

Because Ka is small, assume x < [HA] Ka = 1.4 x = ≈ x2 (0.100 – x) x2 0.100 x = (1.4 x 10-5 )(0.100) = M pH = -log( ) = 2.93 ( << M; the approximation is good) %-ionized = x100 = 1.18 % 0.100 <5% rule can assume [HA]eq~[HA]0

Example 3 – calculating pH from known Kb
What is the pH of a 0.20 M solution of pyridine, C5H5N, in aqueous solution? The Kb for pyridine is 1.4 x 10-9. Initial 0.20 Change -x +x Equilibrium 0.20-x x 2

What is the pH of a 0.20 M solution of pyridine, C5H5N, in aqueous solution? The Kb for pyridine is 1.4 x 10-9. 2

What is the pH of a 0.20 M solution of pyridine, C5H5N, in aqueous solution? The Kb for pyridine is 1.4 x 10-9. Solving for x we get 2

What is the pH of a 0.20 M solution of pyridine, C5H5N, in aqueous solution? The Kb for pyridine is 1.4 x 10-9. Solving for pOH 2

Practice Ka from pH Para-aminobenzoic acid (PABA), HC7H6NO2, is used in some sunscreen agents. A solution is made by dissolving mol of PABA in enough water to make mL of solution. The solution’s pH = What is the Ka for PABA? (1.9x10-5) pH from Ka Barbituric acid (Ka=1.1x10-4) is used in the manufacture of some sedatives. What is the pH for a M solution of barbituric acid? (2.07)

Relationship between Ka and Kb values
For an acid-base conjugate pair: HA and A- [HA][OH-] [A-] Ka x Kb = [H3O+][A-] [HA] = [H3O+][OH-] = Kw Ka Kb When two reactions are added, their equilibrium constants are multiplied (chapter 14).

Relationship between Ka and Kb values
Phenol, C6H5OH, is a weak acid, Ka = 1.3 x at 25°C. Calculate Kb for the phenolate ion C6H5O- Ka x Kb = 1.0 x 10-14 Kb = = 7.7 x 10-5 1.0 x 10-14 1.3 x 10-10

Acid Base Neutralization Reactions
Salt = ionic compound formed in acid + base reaction HX(aq) + MOH(aq) MX(aq) + H2O(l) acid base salt Strong acid + strong base form neutral salts. HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l) H+(aq) + Cl-(aq) + Na+(aq) + OH-(aq) Na+(aq) + Cl-(aq) + H2O(l) Net ionic equation: H+(aq) + OH-(aq) H2O(l) H3O+(aq) + OH-(aq) H2O(l)

Salts of Strong Bases and Strong Acids
H3O+(aq) + OH-(aq) H2O(l) + H2O(l) Keq = 1/Kw = 1x1014 strongacid strong base weak base weakacid 100% H3O+(aq) + Cl-(aq) + Na+(aq) + OH-(aq) Na+(aq) + Cl-(aq) + 2 H2O(l) Na+ and Cl- are spectator ions. Na+ is the v. weak conjugate acid of a v. strong base. Cl- is the v. weak conjugate base of a v. strong acid. Final solution has pH = 7.

Salts of Strong Bases and Weak Acids
Weak acid + strong base form basic salts. CH3COOH(aq) + NaOH(aq) CH3COONa(aq) + H2O(l) CH3COOH(aq) + Na+(aq) + OH-(aq) Na+(aq) + CH3COO-(aq) + H2O 100% Net ionic: CH3COOH(aq) + OH-(aq) CH3COO- + H2O(l) final solution

Salts of Strong Bases and Weak Acids
100% CH3COOH(aq) + OH-(aq) CH3COO-(aq) + H2O(l) base acid acetate is a weak base (much stronger than water) solution is basic (pH > 7.0). CH3COO-(aq) + H2O(l) CH3COOH(aq) + OH-(aq) Calculate pH of solution from Kb of CH3COO-. A hydrolysis reaction – water is broken apart.

pH of a Salt Solution For a weak acid + strong base, pH depends on Kb Larger Kb = stronger base Example What is the pH of a 1.50 M aqueous solution of Na2CO3? Kb (CO32-) = 2.1 x 10-4. (or equal volumes of 1.5M H2CO3 and 3.0 M NaOH are mixed, what is the pH of the mixture?)

pH of a Salt Solution Na+ is the conjugate acid of NaOH (strong base).
What is the pH of a 1.50 M aqueous solution of Na2CO3? Kb = 2.1 x 10-4 Na+ is the conjugate acid of NaOH (strong base). It remains 100% ionized Na+ has no effect on pH. CO32- is the conjugate base of HCO3- (weak acid) Most CO32- ions undergo hydrolysis Generates HCO3- and OH- Changes the pH CO32-(aq) + H2O(l) HCO3-(aq) + OH-(aq)

pH of a Salt Solution Kb = 2.1 x 10-4 = = ≈ [HCO3-][OH-] [CO32-] x2
What is the pH of a 1.50 M aqueous solution of Na2CO3? Kb = 2.1 x 10-4 CO32-(aq) + H2O(l) HCO3-(aq) + OH-(aq) [ ]initial change x x x [ ]equil x x x Kb = 2.1 x 10-4 = = ≈ [HCO3-][OH-] [CO32-] x2 (1.50 – x) 1.50 x = 1.77 x 10-2 pOH = −log(1.77 x 10-2) = 1.75 pH = = 12.25

Salts of Weak Bases and Strong Acids
NH3(aq) + HCl(aq) NH4+(aq) + Cl-(aq) NH3(aq) + H3O+(aq) + Cl-(aq) NH4+(aq) + Cl-(aq) + H2O(l) Net ionic: NH3(aq) + H3O+(aq) NH4+(aq) + H2O(l) The products of the neutralization reaction form an acidic solution: NH4+(aq) + H2O(l) NH3(aq) + H3O+(aq) The final pH depends on the Ka , larger Ka = more acidic

pH of a Salt Solution Find the pH of a M NH4Br. Is this solution acidic, basic or neutral? Ka(NH4+) = 5.6 x The solution contains NH4+ ions and Br- ions. NH4+ is the conjugate acid of NH3 (a weak base). It will partially ionize in solution giving NH3 and H3O+ Forms an acidic solution. Br- is the conjugate base of HBr (a strong acid) Br- ions stay fully ionized Does not change the pH.

pH of a Salt Solution Ka = 5.6 x 10-10 = = ≈ [NH3][H3O+] [NH4+] x2
pH of a M aq. solution of NH4Br? Is this acidic, basic or neutral? Ka(NH4+) = 5.6 x NH4+(aq) + H2O(l) NH3(aq) + H3O+(aq) [ ]initial change x x x [ ]equil x x x Ka = 5.6 x = = ≈ [NH3][H3O+] [NH4+] x2 (0.132 – x) 0.132 x = 8.57 x pH = −log(8.57 x 10-6) = 5.07 Acidic!

Salts of Weak Bases and Weak Acids
The most difficult. Consider qualitative results. e.g. NH3 (aq) + HF(aq) NH4+ (aq) + F- (aq) NH4+(aq) + H2O(l) NH3(aq) + H3O+(aq) F-(aq) + H2O(l) HF(aq) OH-(aq) Ka(NH4+) = 5.6 x ; Kb(F-) = 1.4 x 10-11 Ka (weak acid) > Kb (weak base) The ammonium reaction is more favorable. The acid wins! The solution will be acidic!

Acids, Bases and Salts Strong acid + strong base → salt solution, pH = 7 Strong acid + weak base → salt solution, pH < 7 Weak acid + strong base → salt solution, pH > 7 Weak acid + weak base → salt solution, pH = ? Need K’s “strongest wins”

Neutral Ions in Water (Table 16.5)
Anions Cl- NO3- Br- ClO4- I- Cations Li+ Ca+2 Na+ Sr+2 K+ Ba+2

Example What is the pH of a 0.10 M NaCN solution at 25 oC? The Kb for CN- is 2.5 x 10-5. Sodium cyanide gives Na+ ions and CN- ions in solution. Only the CN- ion hydrolyzes. pH =11.2 2

Example What is the pH of a 0.15 M NH4NO3 solution at 25 oC? The Ka for NH4+ is 5.6 x NH4NO3 gives NH4+ ions and NO3- ions in solution. Only the NH4+ ion hydrolyzes. 2

The Common Ion Effect The common-ion effect is the shift in an ionic equilibrium caused by the addition of a solute that provides an ion common to the equilibrium. Consider a solution of acetic acid (HC2H3O2), in which you have the following equilibrium. 2

The Common Ion Effect If we were to add NaC2H3O2 to this solution, it would provide C2H3O2- ions which are present on the right side of the equilibrium. The equilibrium composition would shift to the left (LeChatelier Principle) and the degree of ionization of the acetic acid would decrease. This repression of the ionization of acetic acid by sodium acetate is an example of the common-ion effect. 2

Example What is the pH of a solution that is 0.025M in formic acid? The Ka for formic acid is 1.7 x What is the pH of the same solution after M in sodium formate, NaCH2O is added? Initial Change Equilibrium 0.018 -x x x 0.025-x x x From NaCH2O 2

Example pH = -log(H3O+) = -log(2.4 x 10-4) = 3.62 Initial 0.025 0.018
0.018 Change -x +x Equilibrium 0.025-x x 0.018+x x [H3O+]= 2.4 × 10-4 pH = -log(H3O+) = -log(2.4 x 10-4) = 3.62 2

Chemistry: The Molecular Science

Similar presentations

Presentation on theme: "Chemistry: The Molecular Science"— Presentation transcript:

Similar presentations

About project

Feedback

Log in

Auth with social network:

Chemistry: The Molecular Science

Similar presentations

Presentation on theme: "Chemistry: The Molecular Science"— Presentation transcript:

Similar presentations

About project

Feedback