2. Bronsted-Lowry theory An acid is a substance from which a proton (H + ion) can be removed A base is a substance that can remove a proton (H + ion) from an acid An acid-base reaction involves the transfer of a proton

3. Types of Acids Acids may have more than one proton to donate. –Monoprotic acids contain 1 hydrogen ion i.e. HCl (aq) –Diprotic acids contain 2 hydrogen ions, i.e. H 2 SO 4 (aq) –Triprotic acids contain 3 hydrogen ions, i.e. H 3 PO 4(aq)

4. Conjugate acid-base pair Two molecules or ions that are related by the transfer of a proton (H+) are called a conjugate acid-base pair. (conjugate means “linked”)

5. The Bronsted-Lowry concept In this idea, the ionization of an acid by water is just one example of an acid-base reaction. Acids and bases are identified based on whether they donate or accept H +. “Conjugate” acids and bases are found on the products side of the equation. A conjugate base is the same as the starting acid minus H +. + ClH H H O + H H HO + acidbase conjugate acidconjugate base conjugate acid-base pairs

6. Example: HBr(aq) + H 2 0(l) -> H 3 0 + (aq) + Br - (aq) Conjugate base of an acid is the particle that remains when a proton is removed from the acid (in this example: Br - (aq) Conjugate acid of a base is the particle that results when the base receives the proton from the acid (in this example: H 3 0 + (aq)

7. Practice problems Identify the acid, base, conjugate acid, conjugate base, and conjugate acid-base pairs: acidbase conjugate acidconjugate base HC 2 H 3 O 2 (aq) + H 2 O (l)  C 2 H 3 O 2 – (aq) + H 3 O + (aq) conjugate acid-base pairs acidbase conjugate acidconjugate base OH – (aq) + HCO 3 – (aq)  CO 3 2– (aq) + H 2 O (l) conjugate acid-base pairs

8. Acids An acid is a substance that produces Hydrogen ions, H + (aq) in water (Arrhenius definition). The hydrogen ion (H + ) bonds with a water molecule to make a hydronium ion, H 3 0 + (aq) (H + -> H 2 0 = H 3 O + )

9. Strong acids A strong acid is an acid, which ionizes extremely well. Essentially all of the acid molecules are ionized. (lots of H 3 O + ) Example: HCl, HBr, HNO 3, H 2 SO 4, HClO 4 Example: HCl (aq) + H 2 O(l) -> H 3 0 + (aq) + Cl - (aq) H 2 SO 4 (aq) +2H 2 O(l) -> 2H 3 O + + SO 4 2- (aq) ( Ionization equations show how they react with water to make H 3 O + )

10. Weak acid A weak acid is an acid which ionizes poorly. Only a small percentage of the acid molecules actually ionize. (a few H 3 O + ) Example:HC 2 H 3 O 2 HC 2 H 3 O 2(aq) + H 2 O(l)  H 3 O+(aq) + C 2 H 3 0 2 -

11. Avoid confusing terms! Concentrated/Dilute vs Strong/Weak A strong acid such as HCl can be concentrated (12mol/L) or dilute (0.5mol/L) Concentrated or dilute refers to the amount of solute dissolved in the solvent Strong or weak- refers to the degree of dissociation of ions in solutions.

12. Bases A base is a substance that produces hydroxide ions, OH -, in water ( Arrhenius defintion ) A strong base is a base that dissociates 100% in aqueous solution. Example: NaOH(s)  Na + (aq) + OH - (aq) KOH(s) -> K + (aq) + OH - (aq) ( Ionization equations: show how they make ions)

13. Weak base A weak base is a base which ionizes poorly. Only a small percentage of the base molecules actually ionize.(a few OH - Example: NH 3 (aq) + H 2 O(l) -> NH 4 + (aq) + OH - (aq)

14. The pH scale A shorter, more suitable method to express the acidity of a solution. pH stands for potency of hydrogen 0 714 Very acidicneutralvery basic [H 3 O + ]>[OH - ] [H 3 O + ]=[OH - ] [H 3 O + ]<[OH - ]

15. pH = -log[ H 3 O + ] Quantitatively: p (power) represents the negative log H stands for the concentration of hydrogen ions or H 3 0 + ions pH= 2.0 [ H 3 O + ] = 1 x 10 -2 mol/L pH= 3.0 [ H 3 O + ] = 1 x 10 -3 mol/L

16. Question: Write the chemical reaction when lithium hydroxide is mixed with carbonic acid. Step 1: write out the reactants LiOH (aq) + H 2 CO 3 (aq)  Step 2: determine products … H 2 O and Li 2 (CO 3 ) LiOH (aq) + H 2 CO 3 (aq)  Li 2 CO 3 (aq) + H 2 O (l) Step 3: balance the equation 2LiOH (aq) + H 2 CO 3 (aq)  Li 2 CO 3 (aq) + 2H 2 O (l) lithium hydroxide + carbonic acid  lithium carbonate + water Writing neutralization equations When acids and bases are mixed, a salt forms NaOH + HCl  H 2 O + NaCl base + acid  water + salt Ca(OH) 2 + H 2 SO 4  2H 2 O + CaSO 4

17. a)iron(II) hydroxide + phosphoric acid b)Ba(OH) 2 (aq) + HCl(aq) c)calcium hydroxide + nitric acid d)Al(OH) 3 (aq) + H 2 SO 4 (aq) e)ammonium hydroxide + hydrosulfuric acid f)KOH(aq) + HClO 2 (aq) Assignment Write balanced chemical equations for these neutralization reactions. Under each compound give the correct IUPAC name.

18. Sample Problem 1 What volume of 0.250M sulfuric acid, H 2 SO 4 (aq) is needed to react completely with 37.2 mL of 0.650 M potassium hydroxide, KOH (aq)? 1.Step 1. Write a balanced equation for the reaction. H 2 SO 4 (aq) + 2KOH (aq)  2H 2 0 (l) + K 2 SO 4 (aq)

19. Step 2. Using the information provided, calculate the number of moles of base. c =.650 mol/L V= 37.2 mL = 0.0372 L n = cV = (.65 mol/L)(0.0372L) = 0.02418 mol H 2 SO 4 (aq) + 2KOH(aq)  2H 2 0(l) + K 2 SO 4 (aq)

20. Step 3. Determine the number of moles of H 2 SO 4 needed to neutralize the lithium hydroxide. (Use mole ratio, if necessary) =0.02418mol KOH X 1 mol H 2 SO 4 2 mol KOH =0.01209 mol H 2 SO 4

21. Step 4. Find the volume of H 2 SO 4 (aq) based on the moles and concentration of H 2 SO 4 solution. c= n/V V= n/c = 0.01209 mol/0.250M = 0.0484 L = 48.4 mL H 2 SO 4 (aq) + 2KOH(aq)  2H 2 0(l) + K 2 SO 4 (aq)

22. Problem 2: A 25.0 mL sample of hydrochloric acid is titrated with 1.00 mol/L NaOH. The end point is reached when 67.5 mL of base has been added. Calculate the concentration of the acid in mol/L. 1.Step 1. Write a balanced equation for the reaction. HCl(aq) + NaOH(aq)  H 2 0(l) + NaCl(aq)

23. Step 2. Using the information provided, calculate the number of moles of base. c = 1.00 mol/L V= 67.5 mL = 0.0675 L n = cV = (1.00 mol/L)(0.0675L) = 0.0675 mol

24. HCl(aq) + NaOH(aq)  H 2 0(l) + NaCl(aq) Step 3. Determine the number of moles of HCl needed to neutralize the sodium hydroxide. (Use mole ratio, if necessary) In this case n A = n B = 0.0675 mol

25. HCl(aq) + NaOH(aq)  H 2 0(l) + NaCl(aq) Step 4. Find the concentration of HCl(aq) based on the moles and volume of HCl solution. c= n/V = 0.0675 mol/0.0250L = 2.70 mol/L

26. Calculate the % of HCl (as m/m) in the solution, given that its density is 1.100 g/mL 2.70 mol HCl 1.00 L of solution m=nM =(2.70 mol)(36.36g/mol) = 98.4 g m=DV =(1.100g/mL)(1000mL) = 1100 g % = 98.4g x 100 1100g = 8.95%

27. Titration Titrations are a chemical procedure involving the addition of a solution of known concentration. The objective is to determine the point in the reaction when the moles of the known solution is equal to the moles of the unknown solution. The concentration of the unknown solution can be determined by: Concentration unknown =Volume known x Concentration known Volume un known

28. Titration Terminology Titrant-solution of known concentration. It is added slowly using a burette Sample-solution of unknown concentration. It is contained in a flask Primary Standards/Solutions- Solution of know accurate concentrations that are used as titrants.

29. Titration Terminology Indicator-A solution that changes color at the equivalence point. Equivalence Point (Stoichiometric Point)-the amount of titrant required for a complete reaction to occur. (chemically equivalent amounts have reacted). Endpoint-The point in a titration where a measurable change occurs (indicator change, pH change)