Applications of Aqueous Equilibria - General Chemistry | CHEM 142, Study notes of Chemistry

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Chapter 8: Applications of Aqueous Equilibria 8.1 Solutions of Acids or Bases Containing a Common Ion 8.2 Buffered Solutions 8.3 Exact Treatment of Buffered Solutions 8.4 Buffer Capacity 8.5 Titrations and pH Curves 8.6 Acid-Base Indicators 8.7 Titration of Polyprotic Acids 8.8 Solubility Equilibria and The Solubility Product 8.9 Precipitation and Qualitative Analysis 8.10 Complex Ion Equilibria Le Châtelier’s principle for the dissociation equilibrium for HF HF(aq) H+(aq) + F-(aq) Adding any salt with F-: “Common ion effect” Ka = [H+][F-] [HF] Like Example 8.1 (P 288) - II Ka = = = 4.0 x 10-4 [H+] [NO2-] [HNO2] ( x ) ( 1.0 + x ) (1.0 – x ) Clearly, x must be small as compared to 1.0 to get right so small: x (1.0) (1.0) = 4.0 x 10 -4 or x = 4.0 x 10-4 = [H+] Therefore pH = - log [H+] = - log ( 4.0 x 10-4 ) = 3.40 The percent dissociation is: 4.0 x 10-4 1.0 x 100 = 0.040 % Nitrous acid Nitrous acid alone + NaNO2 [H+] 2.0 x 10-2 4.0 x 10-4 pH 1.70 3.40 % Diss 2.0 0.040 Example 8.2 (P289-92) - I A buffered solution contains 0.50 M acetic acid (HC2H3O2, Ka = 1.8 x 10-5) and 0.50 M sodium acetate (NaC2H3O2). Calculate the pH of this solution, and the pH after 0.010 M of solid NaOH is added to 1.0 L of this buffer and to pure water. HC2H3O2 (aq) H+(aq) + C2H3O2 (aq) Ka = 1.8 x 10-5 = [H+] [C2H3O2-] [HC2H3O2] Initial Concentration (mol/L) Eqbm Concentration (mol/L) [HC2H3O2]0 = 0.50 [HC2H3O2] = 0.50 – x [C2H3O2-]0 = 0.50 [C2H3O2-] = 0.50 +x [H+]0 = ~ 0 [H+] = x X mol/L of HC2H3O2 dissociates to reach equilibrium Example 8.2 (P289-92) - II Ka = 1.8 x 10-5 = = = [H+][C2H3O2-] [HC2H3O2] ( x ) ( 0.50 + x) 0.50 - x (x) (0.50) 0.50 x = 1.8 x 10-5 The approximation by the 5% rule is fine: [H+] = x = 1.8 x 10-5 M and pH = 4.74 To calculate the pH and concentrations after adding the base: OH- + HC2H3O2 H2O + C2H3O2- Before reaction: 0.010 mol 0.50 mol - 0.50 mol After rxn. (if 100%): 0.010 – 0.010 0.50 – 0.010 - 0.50 + 0.010 = 0 mol = 0.49 mol = 0.51 mol Note that 0.01 mol of acetic acid has been converted to acetate ion by the addition of the base. ~ Added OH- ions are not allowed to accumulate but are replaced by A- : OH- + HA → H2O + A- HA H+ + A- Ka = ← original [HA] and [A-] both [H+] = Ka [HA]/[A-] >> added [OH-] [H+][A-] [HA] How Does a Buffer Work? When the OH- is added, the concentrations of HA and A- change, but only by small amounts. Under these conditions the [HA]/[A-] ratio and thus the [H+] stay virtually constant. The Effect of Added [Acetate-] on pH of an Acetic Acid Buffer [CH3COOH] [CH3COO-]added % Dissociation* pH 0.10 0.00 1.3 2.89 0.10 0.050 0.036 4.44 0.10 0.10 0.018 4.74 0.10 0.15 0.012 4.92 * % Dissociation = x 100 [CH3COOH]dissoc [CH3COOH]init How a Buffer Works–II Let’s consider a buffer made by placing 0.25 mol of acetic acid and 0.25 mol of sodium acetate per liter of solution. What is the pH of the buffer? And what will be the pH of 100.00 mL of the buffer before and after 1.00 mL of concentrated HCl (12 M) is added to the buffer? What will be the pH of 100.00 mL of pure water if the same acid is added? [H3O+] = Ka x = 1.8 x 10-5 x = 1.8 x 10-5 [CH3COOH] [CH3COO-] (0.25) (0.25) pH = -log[H3O+] = -log(1.8 x 10-5) = pH = 4.74 Before acid added! 1.00 mL conc. HCl 1.00 mL x 12 mol/L = 0.012 mol H3O+ If added to 100.00 mL of water : 0.012 mol H3O+ 0.10100 L soln. = 0.120 M H3O+ pH = -log(0.12 M) pH = 0.92 Without buffer! How a Buffer Works–III After acid is added: (Assume first 100% completion) Conc. (M) CH3COOH(aq) + H2O(aq) CH3COO- + H3O+ Initial 0.250 ---- 0.250 0.12 If 100% rxn 0.37 ---- 0.13 0 Solving for the quantity ionized (at real equilibrium): Conc. (M) CH3COOH(aq) + H2O(aq) CH3COO- + H3O+ Initial 0.37 ---- 0.13 0 Change -x ---- +x +x Equilibrium 0.37 - x ---- 0.13 + x x [CH3COOH] [CH3COO-] [H3O+] = Ka x = 1.8 x 10-5 x = 5.1 x 10-5 (0.37) (0.13) Assuming: 0.37 - x = 0.37 & 0.13 + x = 0.13 pH = -log(5.1 x 10-5) = 4.3 After the acid is added! (4.7 before) How Does a Buffer Work? Add a strong acid to a weak acid / salt buffer and see what happens: H+ + A- HA Original buffer pH Final pH of buffer close to original- Added H+ ions Replaced by HA Ka = [H+] [A-] [HA] [H +] = Ka [HA] [A-] How a Buffer Works Buffer after addition Buffer with equal Buffer after addition of H,0* concentrations of of OH” H,0* — conjugate acid and base H,O + CH,COOH ~ H,0* + CH,COO™ CH,COOH + OH~ -» CH,COO~+H,O [H+] = x [CH,COOH] @ [CH,COO-] The Henderson-Hasselbalch Equation Take the equilibrium ionization of a weak acid: HA(aq) + H2O(aq) = H3O+(aq) + A-(aq) Ka = [H3O+] [A-] [HA] Solving for the hydronium ion concentration gives: [H3O+] = Ka x [HA] [A-] Taking the negative logarithm of both sides: -log[H3O +] = -log Ka - log( )[HA] [A-] pH = pKa + log( )[A-] [HA] Generalizing for any conjugate acid-base pair : pH = pKa + log ( )[base][acid] Henderson-Hasselbalchequation The Relation Between Buffer Capacity and pH Change ay ak Oo 0.30 0.10 Concentration (M) of buffer components 0.030 (initial pH = 4.74) 4.8 pH | | T | T | TT TT | T | T 4.9 5.0 5.1 Preparing a Basic Buffer Problem: The ammonia-ammonium ion buffer has a pH of about 9.2 and can be used to keep solutions in the basic pH range. What mass of ammonium chloride must be added to 400.00 mL of a 3.00 M ammonia solution to prepare a buffer ? Plan: The conjugate pair is the ammonia-ammonium ion pair which has an equilibrium constant Kb = 1.8 x 10 -5. The reaction equation with water can be written along with the Kb expression, since we want to add sufficient ammonium ion to equal the aqueous ammonia concentration. Solution: The reaction for the ammonia-ammonium ion buffer is: NH3 (aq) + H2O(l) NH4+(aq) + OH-(aq) Want: [NH4+] = [NH3] [NH3] = 3.00 mol x 0.400 L = 1.20 mol NH4Cl = 53.49 g/mol Therefore mass =NH4Cl = 1.20 mol x 53.49g/mol mass = 64.2 g NH4Cl(s) L Like Example 8.3 (P 285-7) -I Problem: Instructions for making a buffer say to mix 60.0 ml of 0.100 M NH3 with 40.0 ml of 0.100 M NH4Cl. What is the pH of this buffer? The combined volume is 60.0 ml + 40.0 ml = 100.0 ml Moles of Ammonia = VolNH3 x MNH3 = 0.060 L x 0.100 M = 0.0060 mol Moles of Ammonium ion = VolNH4Cl x MNH4Cl = 0.040 L x 0.100 M = = 0.0040 mol [NH3] = = 0.060 M ; [NH4+] = = 0.040 M 0.0060 mol 0.100 L 0.0040 mol 0.100 L Concentration (M) NH3 (aq) + H2O(l) NH4+(aq) + OH-(aq) Initial 0.060 0.040 0 Change -x +x +x Equilibrium 0.060 – x 0.040 – x x Like Example 8.3 (P 295-7) - II Substituting into the equation for Kb: Kb = = 1.8 x 10-5 = [NH4+] [OH-] [NH3] (0.040 + x) (x) (0.060 – x) Assume : 0.060 – x = 0.060 ; 0.040 + x = 0.040~~ Kb = 1.8 x 10-5 = x = 2.7 x 10-5 0.040 (x) 0.060 Check assumptions: 0.040 + 0.000027 = 0.040 or 0.068% 0.060 – 0.000027 = 0.060 or 0.045% [OH-] = 2.7 x 10-5 ; pOH = - log[OH-] = - log (2.7 x 10-5) = 4.57 pOH = 4.57 pH = 14.00 – pOH = 14.00 – 4.57 = 9.43 Exact Treatment of Buffer Solutions Only important: when pKa near 7 and [HA] + [A-] <10-5 M So, read this section so you know the idea and where to find it, but it will not be on the final. Skim next 4 slides. Exact Treatment of Buffer Solutions We can use several relationships to calculate the exact solution to buffered solution problems: Charge – balance equation: [Na+] + [H+] = [A-] + [OH-] Material – Balance equation: [A-]0 + [HA]0 = [HA] + [A-] Since [A-]0 = [Na+] and Kw = [OH-][H+] , we can rewrite the charge balance equation, and solve for [A-] : [A-] = [A-]0 + [H+]2 – Kw [H+] From the mass balance equation solved for [HA] we get: [HA] = [A-]0 + [HA]0 – [A-] Substituting the expression for [A-], and substituting into the Ka expression for HA we obtain: Ka = = [H+][A-] [HA] [H+]{ [A-]0 + } [H+]2 – Kw [H+] [H+]2 – Kw [H+] [HA]0 - Example 8.4 (P 299) - I Calculate the pH of a buffered solution containing 3.0 x 10-4 M HOCl (Ka = 3.5 x 10-8) and 1.0 x 10-4 M NaOCl. Ka = = 3.5 x 10-8 [H+] [OCl-] [HOCl] Let x = [H+] then: [OCl-] = 1.0 x 10-4 + x [HOCl] = 3.0 x 10-4 - x 3.5 x 10-8 = =[H +] [OCl-] [HOCl] (x)(1.0 x 10-4 + x) (3.0 x 10-4 – x ) Assuming x is small compared to 1.0 x 10-4 and solving for x we have: [H+] = x = = 1.05 x 10-7 M = 1.1 x 10-7 M1.05 x 10 -11 1.0 x 10-4 Since this is close to that of water we must use the equation that uses water, and takes it’s ionization into account. Example 8.5 (P 300) - I Calculate the change in pH that occurs when 0.010 mol of gaseous HCl is added to 1.0 L of each of the following solutions: Solution A: 5.00 M HC2H3O2 and 5.00 M NaC2H3O2 Solution B: 0.050 M HC2H3O2 and 0.0500 M NaC2H3O2 For Acetic acid, Ka = 1.8 x 10-5 Use the Henderson-Hasselbalch equation for initial pH: pH = pKa + log{ } [C2H3O2-] [H C2H3O2] Since [C2H3O2-] = [H C2H3O2] The equation becomes: pH = pKa + log (1) = pKa = -log(1.8 x 10-5) = 4.74 Adding 0.010 mol of HCl will cause a shift in the equilibrium due to: H+(aq) + C2H3O2-(aq) H C2H3O2 (aq) Example 8.5 (P 300) - II For Solution A: H+ + C2H3O2- H C2H3O2 Before reaction 0.010 M 5.00 M 5.00 M After 100% rxn 0 4.99 M 5.01 M Calculate the new pH using the Henderson-Hasselbalch equation: pH = pKa + log ( ) = 4.74 + log ( ) = 4.74 – 0.0017 = 4.74 [C2H3O2-] [H C2H3O2] 4.99 5.01 For Solution B: H+ + C2H3O2- → H C2H3O2 Before reaction 0.010 M 0.050 M 0.050 M After 100% rxn 0 0.040 M 0.060 M The new pH is: pH = 4.74 + log( ) = 4.74 – 0.18 = 4.560.040 0.060 Original solution A and new solution Original solution 1.00 H+ New solution [A-] _ 4.99 added [HA] 5.01 = 0.996 pH and Capacity of Buffered Solutions The pH of a buffered solution is determined: mainly by the pKa of the acid, and much more weakly by the ratio [A-]/[HA]: pH = pKa + log ([base]/[acid]) The capacity of a buffered solution is determined: by the lower magnitude of [HA] or [A-]. Buffers well when: [added base or acid] << [HA] and [A-] and [HA] ≈ [A-] Summary: Characteristics of Buffered Solutions Buffered solutions contain relatively large concentrations of a weak acid and its corresponding weak base. They can involve a weak acid HA and the conjugate base A- or a weak base B and the conjugate acid BH+. When H+ is added to a buffered solution, it reacts essentially to completion with the weak base present: H+ + A- HA or H+ + B BH+ When OH- is added to a buffered solution, it reacts essentially to completion with the weak acid present. OH- + HA A- + H2O or OH- + BH+ B + H2O The pH of the buffered solution is determined by the ratio of the concentrations of the weak base and weak acid. As long as this ratio remains virtually constant, the pH will remain virtually constant. This will be the case as long as the concentrations of the buffering materials (HA and A- or G and BH+) are large compared with the amounts of H+ or OH- added. BUFFERS and The Henderson-Hasselbalch equation: pH = pKa + log ([base]/[acid]) For basic buffers this becomes: pH = (14.00-pKb) + log ([base]/[acid]) This named equation can be used for calculating the pH of buffered solutions of acids or bases, and even after adding strong acid or strong base, provided the conc. is small compared to buffer ingredients. Assume they react 100%. Figure 8.1: The pH curve for the titration of 50.0 ml of Nitric acid with 0.10M NaOH Titration of a strong acid (HNO3) solution by a strong base (NaOH). Equivalence point: mol acid = mol base mol H+ = mol OH- Quantitative Analysis How concentrated is the acid sample? Quantify! pH=7.0 for strong acid / strong base titration Eq. Pt. Why does pH change so rapidly near equivalence point? Titrate 50.0mL 0.200M HNO3 (10.0 mmol H+) with 0.100M NaOH 1) Add 10.0 mL NaOH (1.00 mmol) 1.0 mmol OH- + 10.0 mmol H+ = 9.0 mmol H+ [H+] = 9.0 mmol/(50.0mL + 10.0 mL)=0.15M pH = 0.82 2) Add 2 mL more NaOH 1.2 mmol OH- + 10.0 mmol H+ = 8.8 mmol H+ [H+] = 8.8 mmol/(50.0mL + 12.0 mL)= 0.14M pH = 0.85 Change in pH = 0.03 Strong Acid-Base Titration Curve Volume of NaOH added (mL) pH 00.00 1.00 10.00 1.22 20.00 1.48 30.00 1.85 35.00 2.18 39.00 2.89 39.50 3.20 39.75 3.50 39.90 3.90 39.95 4.20 39.99 4.90 40.01 9.40 40.05 9.80 40.10 10.40 40.25 10.50 40.50 10.79 41.00 11.09 45.00 11.76 50.00 12.05 60.00 12.30 70.00 12.43 80.00 12.52 pH 14- 124 10- Titration of 40.00 mL of 0.1000 M HCI with 0.1000 M NaOH } | Phenolphthalein pH = 7.00 at equivalence point T T T T T T T 1 10 20 30 40 50 60 70 80 Volume of NaOH added (mL)  TABLE 8.3 Selected pH Values Near the Equivalence Point in the Titration of 100.0 mL of 0.100 M HCl with 0.100 M NaOH NaOH Added (mL) pH oo) 5.3 100.00 7.0 100.01 8.7 Figure 8.2: The pH curve for the titration of 100.0 ml of 0.50 M NaOH with 0.10 M HCl. Weak Acid- Titration of 40.00 mL of 0.1000 7 HPr with 0.1000 M NaOH Strong Base “ Titration 12 Curve 10 Phenolphthalein . ‘\ = pK, of HPr = 4.89 EZ, pH = 8.80 at equivalence point [HPr]=[Pr] Methyl red Strong acid-strong base curve 0 I T T T T I T ! 10 20 30 40 50 60 70 80 Volume of NaOH added (mL) Treat the stoichiometry Calculating the pH During a Weak Acid-Strong Base Titration–I Problem: Calculate the pH during the titration of 20.00 mL of 0.250 M nitrous acid (HNO2; Ka = 4.5 x 10-4) after adding different volumes of 0.150 M NaOH : (a) 0.00 mL (b) 15.00 mL (c) 20.00 mL (d) 35.00 mL. Plan: (a) We just calculate the pH of a weak acid. (b)-(d) We calculate the amounts of acid remaining after the reaction with the base, and the anion concentration, and plug these into the Henderson-Hasselbalch eq. Solution: HNO2 (aq) + NaOH(aq) H2O(l) + NaNO2 (aq) HNO2 (aq) + H2O(l) H3O+(aq) + NO2-(aq) (a) Ka = = = 4.5 x 10-4 [H3O+] [NO2-] [HNO2] x (x) 0.250 M x2 = 1.125x10-4 x = 1.061x10-2 pH = -log(1.061 x 10-2) = 2.000 - 0.0257 = 1.97 no base added Calculating the pH During a Weak Acid-Strong Base Titration–IV (d) continued Since all of the HNO2 has been neutralized, we only have to look at the concentration of hydroxide ion in the total volume of the solution to calculate the pH of the resultant solution. combined volume = 20.00 mL + 35.00 mL = 55.00 mL [OH-] = = 0.004545 M0.000250 mol OH - 0.05500 L [H3O+] = = = 2.20 x 10-12 Kw [OH-] 1 x 10-14 0.004545 pH = -log (2.200x 10-12) = 11.66 when all of the acid neutralized, and there is an excess of NaOH Figure 8.4: The pH curves for the titrations of 50.0 mL of 0.10 M acids Summary: Titration Curve Calculations A Stoichiometry problem. The reaction of hydroxide ion with the weak acid is assumed to run to completion, and the concentrations of the acid remaining and the conjugate base formed are determined. An equilibrium problem. The position of the weak acid equilibrium is determined, and the pH is calculated. weak Base: ‘a satretbeniot 20,00 mil ot 2-120 ENE rong Aci Titration 12} — INHy]=INH,*) Curve Ph = 5.27 at equivalence point 0 T T T T T T 1 10 20 30 40 50 #60 70 £80 Volume of HCI added (mL) Figure 8.8: The useful pH ranges for several common indicators Figure 8.9: pH curve of 0.10 M HCI being titrated with 0.10 M NaOH Figure 8.11: A summary of the important equilibria at various points in the titration of a triprotic acid TABLE 8.4 A Summary of Various Points in the Titration of a Triprotic Acid Major Species Equilibrium Expression Point in the Titration Present Used to Obtain the pH H*|[HoA No base added H;A, HO eel Base added Before the first equivalence point At the first equivalence point Berween the first and second equivalence points At the second equivalence point Between the second and third equivalence points At the third equivalence point Bevond the third equivalence point H3A, H2A~, H20 H,A ,H,0 H»A~, HA?~, HO HA?-, H,0 HA?~, A?-, H,0 Ae”, H2O A®-, OH”, H30 “1 ~[H:A] K= [H* JHA" | “1 [HA] See the following discussion x = 1 [HA*] [H)A"] See the following discussion = HOE 2 (HA*] y= Ku Ke Ka, _ [HA?|[OH- [Ay] pH determined by excess QH™ DISSOLUTION OF SALTS AND THEIR PRECIPTATION: THE EQUILIBRIUM CONSTANT DESCRIBING SOLUBILITY Table 8.5: Ksp Values at 25 C for Common Ionic Solids Like Example 8.12 (P 330) The Ksp value for the mineral fluorite, CaF2 is 3.4 x 10-11 . Calculate the solubility of fluorite in units of grams per liter. Concentration (M) CaF2 (s) Ca2+(aq) + 2 F-(aq) Initial (unit 0 0 Change activity) +x +2x Equilibrium x 2x Substituting into Ksp: [Ca2+][F-]2 = Ksp (x) (2x)2 = 3.4 x 10-11 4x3 = 3.4 x 10-11 x = x = 2.0 x 10-4 3.4 x 10-11 4 3 The solubility is 2.0 x 10-4 moles CaF2 per liter of water. To get mass we must multiply by the molar mass of CaF2 (78.1 g/mol). 2.0 x 10-4 mol CaF2 x = 1.6 x 10-2 g CaF2 per L 78.1 g CaF2 1 mol CaF2 Relationship Between Ksp and Solubility at 25oC No. of Ions Formula Cation:Anion Ksp Solubility (M) 2 MgCO3 1:1 3.5 x 10-8 1.9 x 10-4 2 PbSO4 1:1 1.6 x 10-8 1.3 x 10-4 2 BaCrO4 1:1 2.1 x 10-10 1.4 x 10-5 3 Ca(OH)2 1:2 6.5 x 10-6 1.2 x 10-2 3 BaF2 1:2 1.5 x 10-6 7.2 x 10-3 3 CaF2 1:2 3.2 x 10-11 2.0 x 10-4 3 Ag2CrO4 2:1 2.6 x 10-12 8.7 x 10-5 The Effect of a Common Ion on Solubility PbCrO4(s) Pb2+(aq) + CrO42-(aq) PbCrO4(s) Pb2+(aq) + CrO42(aq) added Calculating the Effect of a Common Ion on Solubility Problem: What is the solubility of silver chromate in 0.0600 M silver nitrate solution? Ksp = 2.6 x 10-12 . Plan: From the equation and the ion-product expression for Ag2CrO4, we predict that the addition of silver ion will decrease the solubility. Solution: Writing the equation and ion-product expression: Ag2CrO4 (s) 2 Ag+(aq) + CrO42-(aq) Ksp = [Ag+]2[CrO42-] Concentration (M) Ag2CrO4 (s) 2 Ag+(aq) + CrO42-(aq) Initial --------- 0.0600 0 Change --------- +2x +x Equilibrium --------- 0.0600 + 2x x Assuming that Ksp is small, 0.0600 M + 2x = 0.600 M Ksp = 2.6 x 10-12 = (0.0600)2(x) x = 7.22 x 10-10 M Therefore, the solubility of silver chromate is 7.22 x 10-10 M Predicting the Formation of a Precipitate: Qsp vs. Ksp Qsp = Ksp : When a solution becomes saturated, no more solute will dissolve, and the solution is called “saturated.” There will be no changes that will occur. Qsp > Ksp : Precipitates will form until the solution becomes saturated. Qsp< Ksp : Solution is unsaturated, and no precipitate will form. The solubility produce constant, Ksp, can be compared to the ion-product constant, Qsp to understand the characteristics of a solution with respect to forming a precipitate. Predicting Whether a Precipitate Will Form–II Solution cont. For CrO42- : [0.100 M Na2CrO4] [0.200 L] = 0.0200 mol CrO42- [CrO42-] = = 0.667 M in CrO42- 0.0200 mol CrO42- 0.300 liters Qsp = [Ba2+] [CrO42-] =(0.183 M Ba2+)(0.667 M CrO42-) = 0.121 Since Ksp = 2.1 x 10-10 and Qsp = 0.121, Qsp >> Ksp and so a precipitate will form. Figure 8.13: Separation of common cations by precipitation Solution of Hg,”*, Ag’, Pb?*, Hg?*, Cd?*, Bi?*, Cu?*, Sn**, Co?*, Zn?*, Mn?*, Ni2*, Fe2*, Cr3+, Al3*, Ca?*, Ba2?+, Mg?*, NH,*, Nat, K+ Add HCl(aq) Precipitate of Hg,Cl., Solution of AgCl, PbCl, (Group I) Groups II-V Add HS (aq) Y se ag, Precipitate of HgS, CdS, Solution of Bi,S3, CuS, SnS, (Group II) Groups II-V | Add NaQH(aq) oo Precipitate of CoS, ZnS, MnS, NiS, FeS, Solution of Cr(OH);, Al(OH); (Group Il) Groups IV, V Add Na,CO; (aq) Y Precipitate of CaCO3, BaCOs, Solution of MgCoO,; (Group IV) Group V The world faces many challenges for which good science holds the key! Take on that challenge. 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