Chem week 5 notes
Chem week 5 notes CHEM-10055-003-201580
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This 13 page Class Notes was uploaded by Hanna Notetaker on Wednesday September 30, 2015. The Class Notes belongs to CHEM-10055-003-201580 at Kent State University taught by Bansidhar Datta (P) in Fall 2015. Since its upload, it has received 105 views. For similar materials see MOLECULES OF LIFE in Chemistry at Kent State University.
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Date Created: 09/30/15
Chemistry week 5 9/ 28 • Water is amphiprotic, can donate or accept protons • Water can be acid or base • pH + pOH = 14 + o example: H O + 2H O 2 H3O + OH § arrows going both ways show: • equilibrium • rate of going forward is equal to the rate of reverse • dissolves + – • H O 3 = OH -‐ o pOH = -‐ log (OH ) + -‐7 pH = -‐ log (3 O ) pOH= -‐log 10 -‐7 pH= -‐ log 10 pOH= 7 x log 10 pH = 7 x log 10 pOH= 7 -‐ pH= 7 (pH) 7 + (pOH) 7 = 14 • strong acid = dissociates in water 100% • weak acid= dissociates partly in water + -‐ HA + H 2 3 H O + A + -‐ K equilibrium = 3H O ] [A ] ß product 2HA] [H O] ß reactant K a = dissociation constant of weak acid HA K a = H O3] [A ] [HA] Conjugate acids & bases • Acids with or without protons • Bases with or without hydroxide group • Weak acid (HA) / conjugate base (A -‐) • Base (H2O) / conjugate acid (H3O) • The weaker the acid is, the stronger the conjugate base is • The stronger the acid, the weaker the conjugate base is Conjugate Base The conjugate base is made when the acid gives away a proton 2 OH = conjugate base when H2O = acid because H2O gives a hydrogen away to make OH Conjugate Acid = base w/ hydrogen added to it 3 Properties Acids Physical properties: • Sour • Colorless • Odorless • Usually liquid Chemical reactions with: Metals o Metals that react with an acid form metal hydroxide gas § Example: Na + HCl = NaCl + H 2 Metal oxides o Metal oxides that react with acids make salt and water o Example: K + HBr 2 = KBr + H O 2 4 Metal hydroxides (neutralization reaction) o Metal hydroxides that react with acid make a Salt and water o Example: NaOH + HNO =3 NaNO +3 H 2 Bases • Bitter • Slippery • Colorless • Odorless • Solid Applications of neutralization Reaction o Antacids: o Some bases: CaCO ,3 Mh(OH) , 2Al(OH) , 3nd NaHCO o some examples: milk of magnesia, Maalox mylanta, tums o stomachs contains HCl, if there is too much, heartburn can occur, so antacids are taken neutralize o example: CaCo +3 HCl = CaCl2 + 2 O 2 + H O § HCl = in stomach § CaCO3 = tums o balances out/ neutralizes acid in stomach to relieve o NaHCO =3 Sodium Carbonate = baking soda = very strong Applications of Acids: o Vinegar: Acetic acid, CH COOH, used in cooking, preservatives 3 o Boric acid: H BO , antiseptic, for eyes especially 3 3 5 o Phosphoric acid: H PO , in 3io m4cromolecules (proteins, sugars, lipids, nucleic acids) o Hydrochloric Acid: HCl, used in stomachs to digest food Applications of Bases: o Ammonia: NH , fertilizer3, urea, cleaning revives fainted people o Magnesium Hydroxide: MgOH , laxative (8%), “milk2 of magnesia”, metal plant waste, plastics-‐ flame retardant Buffers= a solution that has a weak acid and its conjugate base, resists change when a little amount of acid or base is added to it Buffer capacity= strength of buffer H O2 =least buffer capacity HA/A ratio defines the capacity of the buffer + -‐ Equilibrium-‐ HA H A Henderson Hasselbach equation: HA + H O2 3 H O + A pH = pKa + log A-‐ HA When pH is almost equal to pKa, it is a good buffer 9/30 Strong acids = no Ka (they dissociate completely) Weak acid = has a -‐ K a= (H O)3 (A ) (HA) 6 Acids: -‐5 CH COOH3 Acetic Acid Ka= 1.8 x 10 weaker -‐4 HCOOH Formic Acid Ka= 1.8 x 10 weak H CO Carbonic Acid Ka= 4.3 x 10 -‐7 weakest 2 3 • The lower the value of Ka, the weaker the acid • The higher the pKa, the weaker the acid pKa = -‐log Ka HA + H O 2 H O 3 + A -‐ By switching ( ) it shows that they balance H2O + H2O H3O + OH-‐ K equilibrium = (H3O+) (OH-‐) (H2O) -‐3 + -‐ -‐14 H2O + acid è 10 M [H3O ] [OH ] = 10 -‐3 -‐14 10 x [OH]= 10 -‐14 -‐11 or [OH-‐] = 10 =10 -‐3 10 7 Weak Acid [HA] HA + H2O H3O + A -‐ pH= pKa +log A-‐ HA 1. If HA = 10 [A-‐] then pH= pKa = A = pKa +log(10 ) -‐1 -‐ 10(A ) -‐ 2. if A = 10 [HA] then pH= pKa +log 10 HA =pKa +log10 HA = pKa +1 • Water = BAD BUFFER • Weak acids = GOOD BUFFER • If pH = pKa of HA = best buffer Buffers in the Human Body -‐ OPEN SYSTEM (HCO / CO (d) )3 2 • Open to the environment and physical conditions • HCO = A-‐ CO2= HA d= dissolved • CO2 gas goes to the lungs, its dissolved by alveoli , water& enzyme à H2CO 3 o CO (2) à CO (d)2 2 CO2 (d) + H O 2 (l) 3 à H CO (l) o Equal to: H CO 2 + 3 H O 2 H O3 + HCO 3 -‐ o (l) = liquid (g) = g -‐ o pH= pKa overall +log [HCO ] 3 2 CO 8 o pH= pKa overall +log [HCO ]3 = 6.1 +log (HCO3) 2 C 2 [CO ][HA] 2-‐ -‐ CLOSED SYSTEM (HPO / H 4PO ) 2 4 • Inside cells (enzymatic functions) • PO àH PO 4 3 4 + H 3O +H O4 2 H P2 +H O4 3 pKa = 2.12 + H 3O +H O4 2 H P2 +H O4 3 pKa = 7.2 best ot of 2 equations b/c 7.2 is close to 7.4 (7.4 = pH of blood) + H 3O +H O4 2 H P2 +H O4 3 pKa = 12.1 PLASMA IN OUR BLOOD • pH of blood = 7.4 -‐ • pH= 6.1 +log HCO 3 all together = 7.4 2 CO • conditions: o acidosis § if CO2 in blood is HIGH § pH is less than 6.1 § metabolic acidosis § respiratory acidosis= 9 • CO2 cant get out, too much of the CO2 in the blood, windpipe collapse= Asthma patient o Alkalosis § if CO2 dissolves in blood &there is a LOW amount of plasma § pH = greater than 7.5 § respiratory= • from high fever, bleeding, drug use • metabolic= o enzymes Indicators of pH • Larger molecular compounds change color at varying pH levels o Example: paper that changes color at different pH levels, is the pH level is less than 7 the litmus paper turns red, if pH is bigger than 7 it turns blue § Example: pregnancy test 10 Naming and Molecular Compounds Compounds w/ non metals and metalloids • Number of electrons on the outer shell of non metals and metalloids • Non metals + non metals make molecular compounds • Non metals= C, H, O, N, F, P, S, Cl, Se, Br, I • Metalloids= B, Si, As, Sb, Ge, Te • PIC H has one valence electron that will for bonds H is a atom, 2 makes a molecule 11 12 13