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Chapter 10 Acids and Bases

Chapter 10 Acids and Bases. Acids produce H + ions in water H 2 O H Cl( g ) H+ ( aq ) + Cl  ( aq ) they are electrolytes have a sour taste turn litmus red neutralize bases.

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Chapter 10 Acids and Bases

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  1. Chapter 10 Acids and Bases

  2. Acids produce H+ ions in water H2O HCl(g) H+(aq) + Cl(aq) they are electrolytes have a sour taste turn litmus red neutralize bases

  3. Some acids like sulfuric and phosphoric release more than 1 H+ in water; other like acetic acid (vinegar) release far less than 1 H+ per molecule

  4. Bases produce OH− ions in water are electrolytes feel soapy and slippery neutralize acids NaOH sodium hydroxide KOH potassium hydroxide sodium and potassium hydroxide release 1 OH- /molecule other bases such as ammonium hydroxide (NH4OH) release far fewer OH- fewer

  5. vinegar C2H4O2 ColaH3PO4 Milk of Magnesia (Mg(OH)2 Tums Ca(OH)2

  6. Strong acids completely ionize (100%) in aqueous solutions. • HCl(g) + H2O(l) H3O+(aq) + Cl−(aq) • Amount of acid added Weak acids dissociate only slightly in water to form a solution of mostly molecules and a few ions. H2CO3(aq) + H2O(l) H3O+(aq) + HCO3−(aq)

  7. NH3(g) + H2O(l) NH4+(aq) + OH−(aq) Windex weak base H2CO3 + OH- HCO3- + H2O CO3= + H3O+ Baking Soda weak base NaOH Na+ + OH- Drano strong base

  8. Water reacts with itself in the following manner: H+ is transferred from one H2O molecule to another ; one water molecule acts as an acid, while another acts as a base H2O + H2O H3O+ + OH− .. .. .. .. H:O: + H:O: H:O:H+ + :O:H− .. .. .. .. H H H water water hydronium hydroxide ion(+) ion(-) The concentration of H3O+ = OH- = 10-7 mols/L

  9. pH The pH of a solution is used to indicate the acidity of a solution; it has values that usually range from 0 to 14; the solution is acidic when the values are less than 7; the solution is neutral with a pH of 7; the solution is basic when the values are greater than 7

  10. How is the numerical value of pH determined? pH = - log[H3O+ concentration]; pOH = -log [OH- concentration] when the H3O+ concentration is expressed in mols/L pH + pOH = 14

  11. Reactions of acids and bases Acid + Base = Salt + Water Mg(OH)2 + HCl (gastric juice) = MgCl2 + H2O Mg(OH)2 + 2HCl (gastric juice) = MgCl2 + 2 H2O CaCO3 + HCl = CaCl2 + H2CO3 = CaCl2 + H2O + CO2 CaCO3 + 2HCl = CaCl2 + 2H2CO3 = CaCl2 + 2H2O + 2CO2 + burp

  12. How does the pH vary if we add NaOH (0.1 mol/L) dropwise to a solution of HCl (0.1 mol/L)? HCl + NaOH = H2O + NaCl pH of resulting solution 0.001 0.01 buffered in this region 0.1

  13. How does the pH vary is we add NaOH (0.1 mol/L) dropwise to a solution of the weak acid acetic acid (0.1 mol/L)? HOAc + NaOH = H2O + NaOAC pH of resulting solution buffered in this entire region

  14. Suppose we have a liter of water and we either add a drop of water containing 10-4 moles of HCl or 10-4 moles of NaOH; What would be the resulting pH assuming no volume change with HCl addition? H3O+ = 10-4 mol/L; pH= 4 What would be the resulting pH assuming no volume change with NaOH addition? OH- = 10-4 mol/L; pOH = 4 pH = 14- pOH = 10 alternatively [H+][OH-] = 1 *10-14; [H+] = 10-14/10-4; [H+] = 10-10; pH = 10

  15. How much Mg(OH)2 would be required to neutralize 100 mL of HCl that is 0.1 M? Mg(OH)2 + HCl = MgCl2 + H2O Mg(OH)2 + 2HCl = MgCl2 + 2 H2O Balanced equation How many moles of HCl are their in 100 mL of 0.1 M HCl ? 0.1 M HCl = 0.1 mol/L; 100 mL = 0.1 L 0.1 mol/L *0.1 L = 0.01 moles of HCl 0.5 Mg(OH)2 + HCl = 0.5 MgCl2 + H2O 0.01 moles of HCl requires 0.005 moles of Mg(OH)2

  16. What is the pH of a vinegar solution that is 0.1 M? HOAc + H2O = H3O+ + OAc- What is the equilibrium expression? [H3O+][OAc-]/[HOAc] = K K = 18*10-6 if we let x = [H3O+]; the X also = [OAc-] x2/[0.1-x] = 18*10-6 lets assume that x is very small in comparison to 0.1 x2 = 1.8 * 10-6; x ≈ 1.3*10-3 pH = 2.74

  17. H2CO3 is a very weak acid; however both hydrogens can be removed in the presence of strong base; the pH of a solution of NaHCO3 is very close to physiological pH; in the presence of an acid the HCO3- ion tends to pick up the proton, thus buffering the solution and preventing the solution to become too acidic. • H+ + HCO3- H2CO3 CO2 + H2O In the presence of a base, the HCO3- ion can lose its proton as H+ and thus neutralize the strong base; thus the HCO3- ion can buffer the solution in both directions HCO3- + OH- CO3-2 + H2O

  18. The pH in living systems is very important. For example the pH of blood is kept at 7.4 and must be maintained within ±0.5 pH units. How is this done? At a pH of 7.4, most CO2 is in the form of HCO3- HCO3- can react with either acid or base HCO3- + H3O+ H2CO3 CO2 + H2O HCO3- + OH- H2O + CO3-2 In this manner, HCO3- stabilizes the pH and does not allow it to become too acidic or to basic; it acts as a buffer

  19. CO2 + H2O = H2CO3 = H+ + HCO3-

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