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Acids & Bases. 1. Properties of Acids and Bases: TasteTouch Reactions with Metals Electrical Conductivity Acidsour looks like water, burns, stings Yes-

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Presentation on theme: "Acids & Bases. 1. Properties of Acids and Bases: TasteTouch Reactions with Metals Electrical Conductivity Acidsour looks like water, burns, stings Yes-"— Presentation transcript:

1 Acids & Bases

2 1. Properties of Acids and Bases: TasteTouch Reactions with Metals Electrical Conductivity Acidsour looks like water, burns, stings Yes- produces H 2 gas electrolyte in solution Basebitter looks like water, feels slippery No Reaction electrolyte in solution

3 2. Indicators: Turn 1 color in an acid and another color in a base. A. Litmus Paper: An aciD turns blue litmus paper reD A Base turns red litmus paper Blue. A Base turns red litmus paper Blue. B. Phenolphthalein: colorless in an acid/pink in a base C. pH paper: range of colors from acid to basic D. pH meter: measures the concentration of H + in solution solution

4 3. Neutralization: A reaction between an acid and base. When an acid and base neutralize, salts form. Acid + Base → Salt + Water Ex) HCl + NaOH → NaCl + HOH

5 4. Arrhenius Definition: A. An acid dissociates in water to produce hydrogen ions, H +. B. A base dissociates in water to produce hydroxide ions, OH -. C. Problems with Definition: Restricts acids and bases to water solutions. Oversimplifies what happens when acids dissolve in water. Does not include certain compounds that have characteristic properties of acids & bases. Ex) NH 3 (ammonia) doesn’t fit

6 5. Bronsted-Lowry Definition: A. An acid is a substance that can donate hydrogen ions. Ex) HCl → H + + Cl - B. A base is a substance that can accept hydrogen ions. Ex) NH 3 + H + → NH 4 + C. Advantages of Bronsted-Lowry Definition Acids and bases are defined independently of how they behave in water. Focuses solely on hydrogen ions. D. Hydrogen ion is the equivalent of a proton. Therefore, acids are often called proton donors and bases are called proton acceptors.

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8 6. Hydronium Ion: A. Hydronium Ion – H 3 O + This is a complex that forms in water. B. To more accurately portray the Bronsted- Lowry, the hydronium ion is used instead of the hydrogen ion. C. Amphoteric: A substance that can act as either an acid or a base. Ex.) Water - can gain or lose a H +

9 7. Conjugate Acid-Base Pairs: A pair of compounds that differ by only one hydrogen ion A. When an acid loses a hydrogen ion, it becomes its conjugate base. B. When an base gains a hydrogen ion, it becomes its conjugate acid.

10 Acid (A), Base (B), Conjugate Acid (CA), Conjugate Base (CB) NH 3 +H 2 O ↔NH 4 + +OH - HCl+H 2 O ↔Cl - + H 3 O + A strong acid will have a weak conjugate base. A strong base will have a weak conjugate acid. B B A A CACB CA

11 STRONG Acid/Base C. A strong acid or base will completely dissociate (break apart) in water. This is represented by a single (  ) arrow. HNO 3  H + + NO 3 - NaOH  Na + + OH -

12 D. A weak acid or base will partially dissociate in water. This is represented by a double (↔) arrow. CH 3 COOH ↔ H + + CH 3 COO - NH 4 OH ↔ NH 4 + + OH - WEAK Acid/Base

13 8. Naming Acids Review: A. Binary – H +one anion Prefix “hydro”+ anion name +“ic”acid Ex) HCl hydrochloric acid Ex) H 3 P hydrophosphoric acid B. Tertiary – H + polyatomic anion no Prefix “hydro” (oxo)end “ate” = “ic” acid end “ite” = “ous” acid Ex) H 2 SO 4 sulfuric acid Ex) H 2 SO 3 sulfurous acid

14 15-2 The Self-ionization of Water and pH 1. Water is amphoteric, it acts as both an acid and a base in the same reaction. Ex) H 2 O (l) + H 2 O (l) ↔ H 3 O + (aq) + OH - (aq) 2. In pure water at 25  C, both hydronium ions and hydroxide ions are found at concentrations of 1 x 10 -7 M. Because they are at equilibrium and you do not include liquid water in the equilibrium expression, K eq or K w (water) can be expressed as follows: A. Formula for K w = [H 3 O + ] [OH - ]

15 1.0 x 10 -14 M = [1.0 x 10 -7 M] [1.0x10 -7 M] 1.0 x 10 -14 M = [H 3 O + ] [OH - ] B. Using K w in calculations: If the concentration of H 3 O + in the blood is 4.0 x 10 -8 M, what is the concentration of OH ­ ions in the blood? Is blood acidic, basic or neutral? K w = [H 3 O + ] [OH - ] 1.0 x 10 -14 M = [ 4.0 x 10 -8 M ] [OH - ] 4.0 x 10 -8 M 4.0 x 10 -8 M 2.5 x 10 -7 M = [OH - ] slightly basic

16 3. The pH scale: A.Used to determine if something is an acid or a base. A way to express H 3 O + concentration based on logarithms. pH changes by a factor of 10. Ex) 10,000 = 10 4 therefore log 10,000 = 4 0.001 = 10 -3 therefore log 0.001 = -3 B. pH 1-6.9: acid pH 7.1-14: base pH of 7.0: neutral

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18 D. pH = -log [H 3 O + ] E.[H 3 O + ] [OH - ] = 1.0 x 10 -14 M F.pH + pOH = 14 H+H+ OH -

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20 H+H+ [H 3 O + ] [OH - ] 1x10 -14. 00000000000001 1x10 0 1 1x10 -13.0000000000001 1x10 -1.1 1x10 -12.000000000001 1x10 -2.01 1x10 -11.000000000011x10 -3.001 1x10 -10.00000000011x10 -4.0001 1x10 -9.0000000011x10 -5.00001 1x10 -8.000000011x10 -6.000001 1x10 -7.00000011x10 -7.0000001 1x10 -6.0000011x10 -8.00000001 1x10 -5.000011x10 -9.000000001 1x10 -4.00011x10 -10.0000000001 1x10 -3.0011x10 -11.00000000001 1x10 -2.011x10 -12.000000000001 1x10 -1.11x10 -13.0000000000001 pH 14 13 12 11 10 9 8 7 6 5 4 3 2 1 EQUAL

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22 H. Significant Digits Rule The number of digits AFTER THE DECIMAL POINT in your answer should be equal to the number of significant digits in your original number. Ex -log[8.7 x 10 -4 M] –Calc Answer = 3.0604807474 –Sig Fig pH = 3.06

23 Example #1: [H 3 O + ] = 7.3 x 10 -5 M What is the pH value? Is it an acid, base or neutral? Which equation should you use? K w = [H 3 O + ] [OH - ] OR pH = -log [H 3 O + ] pH = -log [7.3 x 10 -5 M] pH = 4.14 ACID

24 Example #2: [OH - ] = 5.0 x 10 -2 M What is the pH value? Is it an acid, base or neutral? Which equation should you use? K w = [H 3 O + ] [OH - ] OR pH = -log [H 3 O + ] K w = [H 3 O + ] [OH - ] 1 x 10 -14 M = [H 3 O + ] [5.0 x 10 -2 M] 2.0 x 10 -13 M = [H 3 O + ] pH = -log [2.0 x 10 -13 M] pH = 12.70 basic

25 Ionization of Acids & Bases H 2 SO 4  2 H + + SO 4 -2 –Sulfuric acid donates 2 H + ions per mole –This is called a “ diprotic ” acid H 3 PO 3  –Phosphorous acid donates 3 H + ions per mole. This is called a “ triprotic acid” Ca(OH) 2  –Calcium hydroxide dissociates into 2 OH - ions per mole 3 H + + PO 3 -3 Ca +2 + 2 OH -1

26 15-3 Acid-Base Titration 1. An acid-base titration is a carefully controlled neutralization reaction which can determine the concentration [ ] of an unknown solution. 2. To determine the concentration of an unknown substance, a standard solution is needed. This solution has a known concentration. 3. Titration curve: graph that shows how pH changes during a titration.

27 4. An indicator, usually phenolphthalein, is used in a titration. 5. The point at which enough standard solution is added to neutralize the unknown solution is called the equivalence point. 6. The point at which the indicator changes color is called the endpoint.

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29 7. Therefore: [H + ] = [OH - ] at the equivalence point (which is usually the endpoint) Volume (acid) Conc. (acid) = Volume (base) Conc. (base) V a M a = V b M b (L) x Moles = (L) x Moles Liter moles of H + = moles of OH -

30 V a M a = V b M b Ex #1) Solutions of sodium hydroxide are used to unclog drains. A 43.0 mL volume of sodium hydroxide was titrated with 32.0 mL of 0.100 M HCl. What is the molarity of the sodium hydroxide solution? HCl (a) = NaOH (b) (32.0 mL)(0.100 M HCl) = (M b )(43.0 mL) 43.0 mL 43.0 mL 0.0744 M= M b

31 Ex #2) A volume of 25.0 mL of 0.120 M sulfuric acid neutralizes 40.0 mL of a sodium hydroxide solution. What is the concentration of the sodium hydroxide solution? H 2 SO 4 + 2NaOH  Na 2 SO 4 + 2H-OH Note: sulfuric acid has 2 H + per mole of acid Therefore you need to multiply the acid side by a factor of 2 V a M a x 2 = V b M b (25.0 mL) (0.120 M) x 2 = (40.0 mL) x M b M b = 0.150 M

32 Ex #3) 24.9 mL of 2.88 M calcium hydroxide completely neutralizes 38.9 mL of a hydrobromic acid solution. What is the molarity of the hyrdobromic acid? Ca(OH) 2 + 2HBr  CaBr 2 + 2HOH Note: there are 2 hydroxide ions per mole of base Therefore you need to multiply the base side by a factor of 2 V a M a = V b M b x 2 (38.9 mL) x M a = (24.9 mL) (2.88 M) x 2 M a = 3.69 M


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