1 / 16

Acid-Base Reactions

Acid-Base Reactions. Review. Acids are proton (H +1 ion) donors. H 2 SO 4 (aq) + H 2 O(l)  HSO 4 -1 (aq) + H 3 O +1 (aq) In this reaction, H 2 SO 4 (aq) donates a proton to H 2 O. H 2 SO 4 (aq) is the acid in this reaction. Bases are proton acceptors.

kevin-blake
Download Presentation

Acid-Base Reactions

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Acid-Base Reactions

  2. Review • Acids are proton (H+1 ion) donors. • H2SO4(aq) + H2O(l)  HSO4-1(aq) + H3O+1(aq) • In this reaction, H2SO4(aq) donates a proton to H2O. • H2SO4(aq) is the acid in this reaction. • Bases are proton acceptors. • CH3NH2(aq) + H2O(l)  CH3NH3+1(aq) + OH-1(aq) • In this reaction, CH3NH2(aq) accepts a proton from H2O. • CH3NH2(aq) is the base in this reaction. • A conjugate base is the species formed when an acid gives up one proton. • H2PO4-1 is the conjugate base of H3PO4. • A conjugate acid is the species formed when a base accepts one proton. • NH4+1 is the conjugate acid of NH3.

  3. Acid-Base Reactions • Neutralization reaction • acid + base  salt + water • HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l) • H2SO4(aq) + 2 KOH(aq)  K2SO4(aq) + 2 H2O(l) • HNO3(aq) + NH4OH(aq)  NH4NO3(aq) + H2O(l) • Contrary to what the name suggests, neutralization rxns do not always result in a neutral solution. • Depends on the relative strengths of the acid and base that react.

  4. Acid-Base Reactions • Salt – an ionic compound • Formed during an acid-base neutralization rxn • Cation comes from base. • Anion comes from acid. • NaCl is a salt. • Formed from a rxn between HCl and NaOH • HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l) • Mg(ClO3)2 is a salt. • Formed from a rxn between HClO3 and Mg(OH)2 • 2 HClO3(aq) + Mg(OH)2 Mg(ClO3)2(aq) + 2 H2O(l)

  5. Parent Acids and Parent Bases • To determine the parent acid of a salt: • Identify the anion and its charge. • Add as many H+1 ions to the anion as are indicated by its charge. • EXAMPLES: • What is the parent acid of KI? • The anion is I-1. • The parent acid is HI. • What is the parent acid of CuSO4? • The anion is SO4-2. • The parent acid is H2SO4. • What is the parent acid of Mg3(PO4)2? • The anion is PO4-3. • The parent acid is H3PO4.

  6. Parent Acids and Parent Bases • To determine the parent base of a salt: • Identify the cation and its charge. • Add as many OH-1 ions as are indicated by its charge. • EXAMPLES: • What is the parent base of KCl? • The cation is K+1. • The parent base is KOH. • What is the parent base of Ca(NO3)2? • The cation is Ca+2. • The parent base is Ca(OH)2. • What is the parent base of Fe2(SO4)3? • The cation is Fe+3. • The parent base is Fe(OH)3.

  7. Acid-Base Reactions • Determine the “parent acid” and “parent base” of each salt. • NaNO3 • LiBr • Fe(CN)3 • KC2H3O2 • (NH4)2S • HNO3 + NaOH • HBr + LiOH • HCN + Fe(OH)3 • HC2H3O2 + KOH • H2S + NH4OH

  8. Acidic and Basic Salts • Salts can form acidic, basic, or neutral solutions. • In order to determine what type of solution a salt will form, we must compare the strengths of its parent acid and parent base. • Strong acid + strong base  neutral salt • Strong acid + weak base  acidic salt • Weak acid + strong base  basic salt • Weak acid + weak base  varies

  9. Acidic and Basic Salts • Is a solution of NaCl acidic, basic, or neutral? • The parent acid of NaCl is HCl. • HCl is a strong acid. • The parent base of NaCl is NaOH. • NaOH is a strong base. • Conclusion: NaCl forms a neutral solution.

  10. Acidic and Basic Salts • Is a solution of KF acidic, basic, or neutral? • The parent acid of KF is HF. • HF is a weak acid. • The parent base of KF is KOH. • KOH is a strong base. • Conclusion: KF forms a basic solution.

  11. Acidic and Basic Salts • Is a solution of Fe(NO3)3 acidic, basic, or neutral? • The parent acid of Fe(NO3)3 is HNO3. • HNO3 is a strong acid. • The parent base of Fe(NO3)3 is Fe(OH)3. • Fe(OH)3 is a weak base. • Conclusion: Fe(NO3)3 forms an acidic solution.

  12. Titrations • Acid-base titration = a neutralization reaction done for the purpose of determining the concentration of an acidic or basic solution. • EXAMPLE: A solution of HCl with an unknown concentration is reacted with an NaOH solution of known concentration until the HCl solution is completely neutralized. • The concentration of the HCl solution can be calculated from how much NaOH solution was used in the titration.

  13. Titrations • Titrant – the solution whose concentration is known. • Analyte – the solution whose concentration is unknown. • Equivalence point – the point at which stoichiometrically equal amounts of titrant and analyte have been mixed.

  14. Titrations • Formula for use with acid-base titrations: • a x [A] x Va = b x [B] x Vb • a = number of H+1 ions the acid can donate • [A] = molar concentration of the acid • Va = volume of the acid solution • b = number of H+1 ions the base can accept • [B] = molar concentration of the base • Vb = volume of the base solution

  15. Titrations • 25.0 mL of an HCl soln with unknown concentration are titrated with 0.100 M NaOH. • It takes 32.0 mL of the NaOH soln to neutralize the acid soln. • What was the [acid]? • a x [A] x Va = b x [B] x Vb • 1 x [A] x 25.0 mL = 1 x 0.100 M x 32.0 mL • 25.0 mL x [A] = 3.20 M*mL • [A] = 0.128 M

  16. Titrations • 20.0 mL of an HNO3 soln of unknown concentration is titrated with 0.250 M Ca(OH)2. • It takes 18.1 mL of the Ca(OH)2 soln to neutralize the acid soln. • What was [acid]? • a x [A] x Va = b x [B] x Vb • 1 x [A] x 20.0 mL = 2 x 0.250 M x 18.1 mL • 20.0 mL x [A] = 9.05 M*mL • [A] = 0.453 M

More Related