Lecture 7 Professor Hicks Inorganic Chemistry (CHE152) substances - PDF document

Lecture 7 Professor Hicks Inorganic Chemistry (CHE152) substances that Strong acids Acids release H + ions HCl HNO 3 H 2 SO 4 when dissolved H + Cl - H + NO 3 - 2H + SO 4 2- H + + anion hydrochloric acid nitric acid sulfuric acid Weak acids - HC 2 H 3 O 2 HF H+ H + C 2 H 3 O 2 - H + F - anion acetic acid hydrofluoric acid non-metals form anions metals form cations molecular ionic compounds compounds Acids are molecular compounds all other strong acid dissolve molecular acids HCl  H + + Cl - compounds ~ 100% dissolve dissolve HF ~ 95% H + + F - ~5% separated ions (molecules) (separated ions) Dissolved cations anions molecules (+ ions) (- ions)

- Acids H+ anion • Acids are molecular compound because they can dissolve without dissociating into ions • Ionic compounds must separate into ions to dissolve • Weak acids have a small percentage of molecules separated into H + and an anion , the rest stay together as one particle HF ~ 95% H + and F - 5% Strong acids separate 100% into H+ and anion in water HCl ~ 0 % H + and Cl - ~ 100% Hydrogen ion H+ • has no electrons! • to bond other atom must provide both electrons (lone pair) H H H + + N H H H N curved arrows used to show electron pair movement H 1) start at lone pair H 2) end at electron acceptor NH 3 (aq) + H +  NH 4+ (aq) H H carboxylic acids O O O O • weak acids • most common acids in nature C=O C=O C=O C=O R R R R = C,H H - O O O O C=O C=O C=O C=O R = benzene CH 3 CH 3 benzoic acid + H + food preservative R = CH 3 curved arrows acetic acid 1) start at lone pair 2) end at electron acceptor 5% solution = vinegar

hydronium ion H 3 O + + • H 3 O + = H 2 O + H + • form H + takes in water • reactions of acids in water can be written with H + or H 3 O + acid ionization constant K a • equilibrium constant for acid to hydrolyze • acid + H 2 O  H 3 O + + anion Q = [C 6 H 5 CO 2 - ][H 3 O + ] • larger K a = stronger acid [C 6 H 5 CO 2 H] Benzoic Acid = C 6 H 5 CO 2 H (weak acid) - (aq) + H 3 O + (aq) C 6 H 5 CO 2 H (aq) + H 2 O (l)  C 6 H 5 CO 2 K a = 6.5 x 10 -5 reactants reactants products products weak acids strong acids much less almost 100% Gibbs Free Energy Gibbs Free Energy 100% dissociated dissociated HClO 4 K a = 10 10 ! Q = K a 0 Q 2 x 10 -4 0 Q 10 10 Bases H methane • Bronsted-Lowry base = proton acceptor C H H • proton acceptors must have lone pair no lone pairs not a base H Examples of bases hydroxide ion - ionic compounds with OH- ion any hydroxide O-H cation ion 3 lone pairs + OH - cation H ammonia both proton acceptors = both bases N H - molecular compounds with lone pairs lone pair H

conjugate acid/base pairs HC 2 H 3 O 2 (aq) + NaOH (aq)  NaC 2 H 3 O 2 (aq) + H 2 O (l) acid base conjugate base conjugate acid Whenever an acid-base reaction occurs: 1) the product that is “ acid minus H + ” is called the conjugate base of the acid The product that is “base plus H + “ is called the 2) conjugate acid of the base strength conjugate acids/bases • the weaker an acid • the stronger a base the stronger its the weaker its conjugate base conjugate acid conjugate conjugate acid acid base base Q closer to pure acid Q closer to pure conjugate base Gibbs Free Energy Gibbs Free Energy Benzoic acid perchloric acid K a = 6.5 x 10 -5 K a = 10 10 ! 0 Q 2 x 10 -4 0 Q 10 10

Water autoionizes H 2 O (l)  H + (aq) + OH - (aq) Q = [H + ][OH - ] or written with hydronium ion 2H 2 O (l)  H 3 O + (aq) + OH - (aq) Q = [H 3 O + ][OH - ] • K eq for this reaction called K w • K w = 10 -14 at room temperature • Neutral solution has [H + ] =10 -7 and [OH - ] =10 -7

Water is amphoteric (an acid and a base) water acting as a base H 2 O (l) + HCl (aq)  H 3 O + (aq) + Cl - (aq) Conjugate acid of water H 3 O + = hydronium ion water acting as an acid H 2 O (l) + NH 3 (aq)  OH - (aq) + NH 4 + (aq) Conjugate base of water OH - = hydroxide ion Le Chateliers principle and K w what if neutral water is perturbed by adding an acid or base? is system is at equilibrium? yes b/c Q = K a [H 3 O + ][OH - ] = K w = 10 -14 Q = 10 -7 x 10 -7 = 10 -14 2H 2 O (l)  H 3 O + (aq) + OH - (aq) system finds a new equilibrium [H 3 O + ][OH - ] = K w 10 -7 M 10 -7 M 0.10 * [OH - ] = 10 -14 initial M [OH - ] = 10 -14 / 0.10 + 0.10 M =10 -13 M equilibrium M ? ? Why is [H 3 O + ] ~ 0.10 M? (about) If rxn used no OH - [H 3 O + ] = 0.10 + 10 -7 disturbance = add HNO 3 systems response = 0.1000001 increase H 3 O + to 0.10 M = to decrease H 3 O + If rxn used all OH- [H 3 O + ] = 0.10 + 10 -7 -10 -7 =0.10 Acidic solution has [H 3 O + ] > 10 -7 Approximately = 0.10 Basic solution has [OH - ] > 10 -7 pH • scale of [H + ] concentration • more convenient than scientific notation • pH = -log [H 3 O + ] still not sure? take the log of 10 it should be 1 [H 3 O + ]

pOH • same idea as pH • scale of OH- concentration • pOH = -log [OH - ] [H 3 O + ] pH, pOH, and pK w pH is what most people think in terms of pH + pOH = 14 some problems we get a result [OH-] or pOH Why? use this equation to express it as a pH [H 3 O + ][OH - ] = 10 -14 1) take –log both sides 2) log(A*B) = log(A) + log(B) pH + pOH = -log(10 -14 ) = 14 pH = 14 - pOH pH fun facts! • more bacteria that are not harmful grow in acidic conditions ( acidophilus strains ) • more bacteria that are harmful grow in basic conditions • blood pH about 7.4 • stomach pH 1.5 ! • H + and OH - are catalysts for the reactions that hold together, fats, carbohydrates, and proteins!!!!  control of pH important for life

The Strong Acids HCl, HBr, HI, HNO 3 , HClO 4 , H 2 SO 4 • molarity of a monoprotic strong acid = molarity of [H 3 O + ] • b/c strong acids completely, 100% dissociate For example a 1.0 M solution of HCl has a [H 3 O + ] = 1.0 M

Dissociation of weak acids Calculate [H 3 O + ] at equilibrium for a 0.55 M solution of HF in water. HF (aq) + H 2 O (l)  H 3 O + (aq) + F - (aq) write hydrolysis reaction Acid + H 2 O  H 3 O + + conj base 0.55 - 0 0 initial (M) -x - +x +x change (M) 0.55-x - +x +x equilibrium (M) [H 3 O + ][F-] K a = to solve for x you could use the quadratic equation [HF] a quicker alternative is to look for an approximation look up K a for HF if x is very small compared to 0.55 x 2 x 2 3.5 x 10 -5 = 3.5 x 10 -5 = x = ??? 0.55 0.55-x x = sqrt{ 0.55* 3.5 x 10 -5 } = 0.004387 M x = change in [HF] to reach equilibrium, and final equilibrium molarities [H 3 O + ], [F - ] “x” Percent % dissociated = x 100% initial Dissociated molarity initial HF before molarity dissociation dissolve initial H + (aq) - x x HF (aq) molarity F- (aq) at equilibrium

Dissociation of weak acids HF (aq) + H 2 O (l)  H 3 O + (aq) + F - (aq) 0.55 - 0 0 initial (M) -x +x +x change (M) 0.55-x +x +x equilibrium (M) [H 3 O + ][F-] to solve for x you could use the quadratic equation K a = [HF] a quicker alternative is to look for an approximation if x is very small compared to 0.55 x 2 x 2 3.5 x 10 -5 = 3.5 x 10 -5 = x = ??? 0.55 0.55-x x = sqrt{ 0.55* 3.5 x 10 -5 } = 0.004387 M % dissociated is 0.004387 is small compared to 0.55 M? 5% error is widely accepted in science 0.004387 approximation is valid b/c % x 100% = 0.79% 0.55 dissociated is less than 5% so error could not be larger than 5% The method of 2- (aq) + H 2 O (l)  HCO 3 - (aq) + OH - (aq) CO 3 successive initial (M) 0.010 - 0 0 change (M) -x +x +x approximations equilibrium (M) 0.01 - x +x +x x 2 x 2 1.8 x 10 -4 = % dissociated = 0.00134 x100% = 13.4% 0.010 -x 0.010 - 0.00134 0.010 x = sqrt{ 0.01* 1.8 x 10 -4 } = 0.00134 M more than 5%! approximation not good ! 1) substitute this approximation of x back into equation use the method of successive approximations if % dissociation 2) calculate new approximate value of x is more than 5% x = sqrt { 1.8x10 -4 (0.01 - 0.00134 } an improved approximation of x = 0.00128 repeat steps 1, 2 x = sqrt { 1.8x10 -4 (0.01 - 0.00128 } a more improved approximation of x =0.00125 repeat steps 1, 2 again the approximations have stopped x = sqrt { 1.8x10 -4 (0.01 - 0.00125 } changing in the 5th decimal place so the approximation good to this precision = 0.00125

1) Break into groups of 2-3 - each group will be assigned an acid 2) Determine the pH and % dissociated. assigned acid K a M 1) HClO 2 1.10 x 10 -02 5.0 2) HCHO 2 1.80 x10 -04 9.0E-02 3) C 7 H 6 O 2 6.50 x10 -05 3.0E-02 4) HC 2 H 3 O 2 1.80 x10 -05 1.0E-02 5) HClO 2.90 x10 -08 2.0E-05 6) HCN 4.90 x10 -10 2.5E-07 7) HC 6 H 5 O 1.30 x10 -10 6.0E-08 8) HF 3.50 x10 -04 1.8E-01 9) HNO 2 4.60 x10 -04 2.0E-01 Determine the pH and % dissociated for a 1.5 x 10-4 M solution of acetic acid using the method of successive approximations Le Chateliers Principle and % dissociated HF + H 2 O  F  + H 3 O + K a = 3.5 x 10 -4 10% 40% 1% 30% ~100% 90% 50% 60% 70% 80% at “infinite” dilution percent dissociated molarity declines with dilution disturbance = increase H 2 O (dilute acid) 1.0 M  0.10 M  0.001 M  response = decrease [H 2 O] 0.00010 M  etc