1. Acid-Base Reactions Ch. 15

2. Acid-Base Reactions Neutralization reactions Neutralization reactions – pH is changed Produce a salt and H 2 O Produce a salt and H 2 O – Salts are ionic compounds 2 types of Acids 2 types of Acids – Strong and Weak 2 types of Bases 2 types of Bases – Strong and Weak 4 possible combinations of Acids and Bases 4 possible combinations of Acids and Bases – Strong A + Strong B – Strong A + Weak B – Weak B + Strong B – Weak B + Weak A

3. Strong Acid + Strong Base HCl + NaOH  HCl + NaOH  Double replacement reaction Double replacement reaction Both compounds completely dissociate Both compounds completely dissociate HCl  Cl - + H + NaOH  Na + + OH - Complete equation (aq) can be written: Complete equation (aq) can be written: H + +Cl - +Na + +OH -  Cl - +Na + +H 2 O Spectator Ions Spectator Ions – Ions that do not take part in the reaction

4. Net Ionic Equations 1) Write Complete Ionic Equation All ionic compounds are shown as free ions 2) Remove Spectator Ions Ions not directly evolved in the rxn. 3) Balance the remaining rxn.

5. Strong Acid + Strong Base What is the net ionic equation: What is the net ionic equation: HCl (aq) + NaOH (aq) 1) H + (aq) +Cl - (aq) +Na + (aq) +OH - (aq)  Cl - (aq) +Na + (aq) +H 2 O (aq) 2) H + (aq) +OH - (aq)  H 2 O (l) All strong acid and strong base reactions have this as a base net ionic equation All strong acid and strong base reactions have this as a base net ionic equation KOH (aq) +HNO 3(aq)  KNO 3(aq) +H 2 O (l) K + +OH - +H + +NO 3 -  K + +NO 3 - +H 2 O (l) K + +OH - +H + +NO 3 -  K + +NO 3 - +H 2 O (l) OH - +H + +  H 2 O (l) OH - +H + +  H 2 O (l)

6. Strong Acid + Weak Base HCl (aq) + Al(OH) 3(s) HCl (aq) + Al(OH) 3(s) Weak bases wont completely dissociate Weak bases wont completely dissociate – Cannot write them as ions on reaction side of net ionic equation H + (aq) +Cl - (aq) +Al(OH) 3(s)  Al +3 (aq) +Cl - (aq) +H 2 O (l) H + (aq) +Al(OH) 3(s)  Al +3 (aq) +H 2 O (l) H + (aq) +Al(OH) 3(s)  Al +3 (aq) +H 2 O (l) End solution is slightly acidic End solution is slightly acidic What about NH 3 ? What about NH 3 ? – Considered a weak base but has no OH - – Does not produce water – HCl (aq) +NH 3(aq)  NH 4 Cl (aq) H + (aq) +Cl - (aq) +NH 3(aq)  NH 4 + +Cl - (aq) H + (aq)) +NH 3(aq)  NH 4 + H + (aq)) +NH 3(aq)  NH 4 +

7. Weak Acid + Strong Base HC 2 H 3 O 2(aq) + NaOH (aq) HC 2 H 3 O 2(aq) + NaOH (aq) Weak acid wont completely dissociate Weak acid wont completely dissociate – Wont breakdown into ions on reaction side HC 2 H 3 O 2(aq) + Na + (aq) +OH - (aq)  Na + (aq) +C 2 H 3 O 2 - (aq) + H 2 O (l) HC 2 H 3 O 2(aq) + Na + (aq) +OH - (aq)  Na + (aq) +C 2 H 3 O 2 - (aq) + H 2 O (l) HC 2 H 3 O 2(aq) +OH - (aq)  C 2 H 3 O 2 - (aq) + H 2 O (l) HC 2 H 3 O 2(aq) +OH - (aq)  C 2 H 3 O 2 - (aq) + H 2 O (l) End solution is slightly basic End solution is slightly basic Weak Acid + Weak Base ???Not clear??? Both the acid and base are so unreactive there is little change Both the acid and base are so unreactive there is little change Not common reaction type in nature Not common reaction type in nature

8. Bronsted-Lowery Acids and Bases Acids produce H + ions when added to water Acids produce H + ions when added to water Bases produce OH - ions when added to water Bases produce OH - ions when added to water HCl (aq) +H 2 O (l)  H 3 O + (aq) +Cl - (aq) H 2 O (l) +NH 3(aq)  NH 4 + (aq) +OH - (aq) Water can act as an acid or a base Water can act as an acid or a base Acid= Any compound that releases H + Acid= Any compound that releases H + Base= Any compound that takes H + Base= Any compound that takes H + Conjugate Acid/Base Conjugate Acid/Base – Weak acid or base produced from an acid-base reaction ACID BASEACID BASEConjugate acid Conjugate base

9. Regulating pH Living things interact with acids and bases all the time; their pH must be regulated Living things interact with acids and bases all the time; their pH must be regulated Buffer Buffer – Solution that adjusts to the addition of acids and bases to slowly change the pH – Free OH - and H + ions Weak acic/base + salt of that acid/base Weak acic/base + salt of that acid/base NaOH + HC 2 H 3 O 2  H 2 O + NaC 2 H 3 O 2 NaC 2 H 3 O 2 and HC 2 H 3 O 2 Add a strong base: HC 2 H 3 O 2(aq) +NaOH (aq)  C 2 H 3 O 2 - (aq) +H 2 O (l) HC 2 H 3 O 2(aq) +OH - (aq)  C 2 H 3 O 2 - (aq) +H 2 O (l) Add a strong acid: NaC 2 H 3 O 2(aq) +HCl (aq)  HC 2 H 3 O 2(aq) +NaCl (l) C 2 H 3 O 2 - (aq) +H + (aq) +  HC 2 H 3 O 2(aq)

10. Buffers in the Blood Blood must keep a pH of 7.4 to allow the best exchange of CO 2 and O 2 Blood must keep a pH of 7.4 to allow the best exchange of CO 2 and O 2 Blood buffer is HCO 3 - /H 2 CO 3 Blood buffer is HCO 3 - /H 2 CO 3 Add Base: H 2 CO 3 +OH -  HCO 3 - +H 2 O Add Base: H 2 CO 3 +OH -  HCO 3 - +H 2 O Add Acid: HCO 3 - +H +  H 2 CO 3 Add Acid: HCO 3 - +H +  H 2 CO 3 What happens when you take in too much CO 2 ? – H 2 CO 3 increases making blood more acidic What kind of blood pH results in yawning? – Acidic blood; body needs to release large amount of CO 2 by taking in large amount of O 2

11. Antacids Compound controls acidic pH levels by adding base Compound controls acidic pH levels by adding base 2 types: 2 types: – Hydroxide Antacids Low solubility in water Low solubility in water Release OH - to neutralize H + Release OH - to neutralize H + Milk of Magnesia (Mn(OH) 2 ) Milk of Magnesia (Mn(OH) 2 ) – Carbonate Antacids XCO 3 or XHCO 3 ; react with HCl XCO 3 or XHCO 3 ; react with HCl CaCO 3 + 2HCl  CaCl 2 + H 2 CO 3 CaCO 3 + 2HCl  CaCl 2 + H 2 CO 3 Breaks down to CO 2 and H 2 O

12. Titrations Test to determine the molarity of an acid or a base Test to determine the molarity of an acid or a base – Find the Standard Solution Process: Process: – Standard solution of an acid/base is slowly added to an acid/base of unknown molarity – When the unknown acid/base is neutral, the [H + ]=[OH - How do we know the Standard solution is neutral? How do we know the Standard solution is neutral? pH Indicators pH Indicators – Volume of acid/base used gives us molarity – M A V A = M B V B End point Equivalence point Neutral point

13. Titrations practice If 15.0 mL of 0.50 M NaOH is used to neutralize 25.0 mL of HC 2 H 3 O 2, what is the molarity of the acid solution? NaOH + HC 2 H 3 O 2  H 2 O + NaC 2 H 3 O 2; 1:1 ratio M A V A = M B V B M A V A = M B V B M A = M B V B /V A = (0.50 M)(15.0 ml)/25.0 ml = 0.30 M M A = M B V B /V A = (0.50 M)(15.0 ml)/25.0 ml = 0.30 M If 25.0 mL of a standard 0.05 M HCl solution is required to neutralize 20.0 mL of a solution of Sr(OH) 2, what is the concentration of the base? 2 HCl + Sr(OH) 2 ® SrCl 2 + 2H 2 O M A V A = 2 M B V B M A V A = 2 M B V B M B = M A V A /2V B = (0.05 M)(25.0 ml)/(2)(20.0 ml) = 0.03 M

14. Titrations practice What types of acid-base titrations do these graphs show?

15. Titrations practice Graphs shows titration of 0.5 M NaOH with 50ml of an unknown acid. After titration NaBr salt crystals were isolated from the solution. a) What is the acids used? Is it strong or weak? b) what is the concentration of the acids used? HBr: NaOH + HBr  NaBr + H 2 O; strong M A = M B V B /V A = (0.5M)(50ml)/35ml = 0.7 M