Grace King High School Chemistry Test Review

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1 CHAPTER 19 Acids, Bases & Salts 1. ACIDS Grace King High School Chemistry Test Review UNITS 7 SOLUTIONS &ACIDS & BASES Arrhenius definition of Acid: Contain Hydrogen and produce Hydrogen ion (aka proton), H + in water (aqueous solutions) HCl H + + Cl - Examples: HCl (hydrochloric acid), HNO 3 (Nitric acid), H 2 SO 4 (Sulfuric acid), H 3 PO 4 (Phosphoric acid) CH 3 COOH (acetic acid; vinegar acid), H 2 CO 3 (carbonic acid) etc. Bronsted-Lowry definition: Hydrogen ion donor NH + 4 H + + NH 3 Example: NH + 4 (ammonium ion) Properties: Sour taste, electrolytes-conduct electricity, react with metals to release H 2 gas, turn blue litmus to red, ph of acid solutions < 7.0 Monoprotic/Polyprotic: Monoprotic (release 1 H + in solution) or Polyprotic (release more than 1 H + in solution) Conjugate acid: Formed when a base receives a H + (proton) - Example : HCO 3 + H+ H 2 CO 3 (conjugate base) (proton) (conjugate acid) 2. BASES Arrhenius definition of Base: Produce Hydroxide ion OH - in water (aqueous solutions) Examples: NaOH (sodium hydroxide), Mg(OH) 2 (Magnesium hydroxide), KOH (potassium hydroxide), Bronsted-Lowry definition: Hydrogen ion acceptor NH 3 + H + NH 4 + Example: NH 3 (ammonia) Properties: bitter taste, electrolytes-conduct electricity, slippery to touch, turn red litmus to blue, ph of basic solutions > 7.0 Monobasic/Polybasic: Monobasic (release 1 OH - in solution) or Polybasic (release more than 1 OH - in solution) Conjugate base: Formed when an acid donates a H + (proton) Example : H 2 CO 3 H + + HCO 3 - (conjugate acid) (proton) (conjugate base)

2 3. AMPHOTERIC Definition: A substance that can act as an acid or a base Example: Water HCl + H 2 O H 3 O + + Cl - (water accepts a proton, H +, therefore acts as a base) + NH 3 + H 2 O NH 4 + OH - (water donates a proton, H +, therefore acts as an acid) 4. WATER Concentration of H + and OH - in water [H + ] = 1 x 10-7 M & [OH - ] = 1 x 10-7 M at 25 0 C Water is neutral Because [H + ] = [OH - ] in pure water [ ] means concentration Ion product of Water (Kw) Kw = [H + ] x [OH - ] = 1 x 10-7 M X 1 x 10-7 M = (no units) 5. CONCEPT OF ph & poh Definition of ph ph = -log 10 [H + ] ph of Water ph of water = -log[h + ] = -log (1 x 10-7 ) = -(log 1 + log 10-7 ) = -(0.0 + (-7log10)) = -(-7 x 1) = 7 poh of Water poh of water = -log[oh - ] = -log (1 x 10-7 ) = -(log 1 + log 10-7 ) = -(0.0 + (-7log10)) = -(-7 x 1) = 7 ph + POH of Water ph + poh = = ph and ACIDIC, BASIC & NEUTRAL SOLUTIONS Acidic Solutions [H + ] > [OH - ] Therefore, [H + ] > 1 x 10-7 M and ph < 7.0 Assume that [H + ] = 9.9 x 10-6 M ph = -log [H + ] = -log [9.9 x 10-6 ] = 6.99 Therefore, ph is less than 7 Basic Solutions [OH - ] > [H + ] Therefore, [OH - ] > 1 x 10-7 M and ph > 7.0 Assume that [OH - ] = 9.9 x 10-6 M poh = -log [OH - ] = -log [9.9 x 10-6 ] = 6.99 ph + poh = 14, Therefore, ph = 14 poh = = 7.01, which is greater than 7

3 Neutral Solutions [H + ] = [OH - ] Therefore, ph = poh = 7 ph Scale Ranges from 0-14 ph Measurements ph may be measured using 1. Acid-base indicators or, 2. Using a ph meter Acid-Base Indicators Indicators are chemicals that change color in certain ph ranges. May be used in a liquid form in solutions or in a solid form by adsorbing onto paper. Litmus paper (Red & Blue) litmus is the most commonly used ph indicator to test if a solution is acidic or basic. ph paper will turn red in acidic solutions and blue in basic solutions ph meter Uses a probe and electronics to measure ph. 7. STRONG & WEAK ACIDS Strength of an Acid is an Intensive Property The strength of an acid depends on the nature (intensive) and not on the amount (extensive) Degree of Dissociation The strength of an acid depends on the degree of dissociation, or the extent to which it ionizes in water. The strongest acids dissociate (ionize) completely producing the maximum [H + ] HCl + H 2 O H 3 O + + Cl - 0% 100% The weaker acids do not dissociate (ionize) completely producing less than the maximum [H + ] CH 3 COOH + H 2 O H 3 O + + CH 3 COO - 99% 1% Acid Dissociation Constant (Ka) HA + H 2 O H + + A - K a = [H + ] x [A - ] [HA] The higher the Ka, the stronger the acid. Acid Type Ka pka Very strong acid >0.1 <1 Moderately strong acid Weak acid Very weak acid Extremely weak acid >15

4 8. STRONG & WEAK BASES Strength of a Base is an Intensive Property The strength of a base depends on the nature (intensive) and not on the amount (extensive) Degree of Dissociation The strength of a base depends on the degree of dissociation, or the extent to which it ionizes in water. The strongest bases dissociate (ionize) completely producing the maximum [OH - ] KOH + H 2 O K + + OH - 0% 100% The weaker bases do not dissociate (ionize) completely producing less than the maximum [OH - ] H 2 NNH 2 + H 2 O + H 2 NH 3 + OH - 99% 1% Base Dissociation Constant (Kb) BA + H 2 O BAH + + OH - K b = [BAH + ] x [OH - ] [BA] The higher the Kb, the stronger the base. 9. ACID-BASE REACTIONS Acid-Base Reactions-General Observations The reaction goes to completion when [H + ] = [OH - ] Therefore, these types of reactions are called Neutralization Reactions. Acid-Base Titrations Titration is a method for estimating the concentration of an unknown solution by reacting a known volume of it with a second solution of known concentration and volume Acid-Base titrations involve an acid and a base + an indicator to determine the end point. The unknown may be the acid or the base. For a strong acid and a strong base, the equivalence point is ph 7.0. For a weak acid titrated with a strong base, the equivalence point is >ph 7.0. For a weak base titrated with a strong acid, the equivalence point is <ph 7.0. Base Type Kb pkb Very strong base >0.1 <1 Moderately strong base Weak base Very weak base Extremely weak base >15

5 10. QUANTIFYING ACID-BASE REACTIONS Mole Ratios The equation for a neutralized reaction may be written just like the other types of reactions. The reaction must be balanced because these reactions also obey the Law of Conservation of Mass. The balanced reaction gives us the mole ratios for quantifying reactants and products. Mole Ratios Examples 1 HCl(aq) + NaOH(l) è NaCl (aq) + H 2 O(aq) 1 mole 1 mole 1 mole 1 mole 1 mol HCl 1 mol NaOH 1 mol NaCl 1 mol NaOH 1 mol H 2 O 1 mol NaOH 1 mol HCl 1 mol H 2 O 2 H 2 SO 4 (aq) + 2NaOH(aq) è Na 2 SO 4 (aq) + 2H 2 O(l) 1 mole 2 mole 1 mole 2 mole 1 mol H 2 SO 4 1 mol Na 2 SO 4 2 mol H 2 O 1 mol H 2 SO 4 N 1 V 1 = N 2 V 2 N 1 = Normality of reactant 1 V 1 = Volume of reactant 1 N 2 = Normality of reactant 2 V 2 = Volume of reactant 2 Normality versus Molarity Molarity = moles of solute/l of solution Molarity = equivalent moles of solute/l of solution H 2 SO 4 (aq) + 2NaOH(aq) è Na 2 SO 4 (aq) + 2H 2 O(l) 1 mole 2 mole 1 mole 2 mole For the reaction on the right, Molarity of H 2 SO 4 = 1mol but normality of H 2 SO 4 = 2 mol Molarity of NaOH = 1 mol and normality of NaOH = 1 mol 11. PROBLEM SOLVING Concentrations have to be in molar (M) 1. Calculate ph from [H + ] Ex: [H + ] = 1.6 x 10-2 M, calculate ph. ph = -log [H + ] = -log [1.6 x 10-2 ] = -(log1.6 + (- 2log10)) = -(0.20 2) = -(-1.8) = Calculate ph from [OH - ] Ex: [OH - ] = 3.3 x 10-5 M, calculate ph. poh = -log [OH - ] = -log [3.3 x 10-5 ] = -(log3.3 + (- 5log10)) = -(0.52 5) = -(-4.48) = 4.48 ph + poh = 14 ph = 14 -poh = = Calculate [H + ] from ph Ex: ph = 13.20, find [H + ] -log [H + ] = ph -log [H + ] = 13.2 [H + ] = = x = 6.31 x Calculate the concentration of an unknown acid or base solution (normality = molarity) What volume of 0.1M NaOH is required to completely neutralize 50 ml of 0.25M HCl? N 1 V 1 = N 2 V 2 (acid) N 1 = 0.25M V 1 = 50 ml (base) N 2 = 0.1M V 2 =? ml 0.25M x 50 ml = 0.1M x V 2 V 2 = 0.25 M x 50 ml = 125 ml 0.1 M

6 5. Calculate the concentration of an unknown acid or base solution (normality molarity) What volume of 0.12M HCl is required to completely neutralize 33 ml of 0.23M Mg(OH) 2? N 1 V 1 = N 2 V 2 (base) N 1 = 0.23M x 2 = 0.46M V 1 = 33 ml (acid) N 2 = 0.12M V 2 =? ml 0.46M x 33 ml = 0.12M x V 2 V 2 = 0.46M x 33 ml = ml 0.12M