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Unit Four. Acids-Bases Reactions. Acids & Bases. What causes acid rain? And how can we prevent the damage? Why do Perrier drinking chickens give better eggs than chickens who drink regular water? These are the types of questions we will be able to answer after this unit.
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Unit Four Acids-Bases Reactions
Acids & Bases • What causes acid rain? And how can we prevent the damage? • Why do Perrier drinking chickens give better eggs than chickens who drink regular water? These are the types of questions we will be able to answer after this unit.
First Proposed Theory: Arrhenius Theory of Acids & Bases An acid is any substance which releases ___________ in water - it is any ionic species that start with “____” - it tastes sour, & turns blue litmus paper red eg. HCl, HNO3, H2SO4 An base is any substance which releases ___________ in water - it is any ionic species that ends with “____” - it tastes bitter, feels slippery, & turns red litmus paper blue eg. NaOH, KOH, Ca(OH)2
A ___________ results from the neutralization rxn. of an acid & base Neutralization rxn: Acid + Base → salt + H2O Net Ionic Equation: ________________________ Eg. HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
Q: Predict the products & balance the following Neutralization Reactions H2SO4(aq) + NaOH(aq) → H3PO4(aq) + KOH(aq) → H4P2O7(aq) + NaOH(aq) →
Second Proposed Theory: Bronsted-Lowry Theory of Acids & Bases • more general & includes equilibrium rxns An acid is a substance that DONATES a ___________ (___) An base is a substance that Acceptsa ___________ (___) Eg. NH3 + H2O NH4+ + OH- NH3 gained a proton → NH4+ Thus, NH3 = ___________ H2O lost a proton → OH- Thus, H2O = ___________
Your Turn! CH3COOH + H2O CH3COO- + H3O+ (lost/donated H+) (gained H+) ACID BASE Which reactants acts as an acid & which as a base? • HNO3 + H2O NO3- + H3O+ • HCO3- + SO32- CO32- + HSO3- • HS- + H2PO4- H2S + HPO42- • H3PO4 + CH3COO- H2PO4- + CH3COOH • CO32- + HF HCO3- + F-
Classifying Acids A ___________ protic acid can donate 1 proton. A ___________ protic acid can donate up to 2 proton. A ___________ protic acid can donate up to 3 proton. All acids that can donate more than 1 proton are known as ___________ protic acids.
Amphiprotic Substances Amphiprotic substances can act as an ___________ or a___________. eg. Water NH3 + H2O NH4+ + OH- Base acid CH3COOH + H2O CH3COO- + H3O+ Acid base
Apart from water, amphiprotic substances start with ____ and have a (_______) charge. eg. H2PO4- , HS-, HCO3- Base acid H3PO4 H2PO4- HPO42- Accepts H+ Donates H+
NOTE: In every Bronsted-Lowry rxn. there is an acid and a base on both sides of the eqn. CH3COOH + H2O CH3COO- + H3O+ acidbase base acid
Q: Which do you think exhibit amphiprotic behavior? Se2- HSe- H2Se H3PO4 HPO42- HSO3-
Q: Identify each species as an acid or base. • HF + SO32- F- + HSO3- • H2O + HCO3- H3O+ + CO32- • NO2- + H2O OH- + HNO2 • H2PO- + S2- HS- + HPO42- • N2H5+ + SO42- N2H4 + HSO4-
Conjugate Acid & Bases A “conjugate” acid-base pair is a pair of chemical species that only differ by ONE proton. Eg. NH4+ + H2O NH3 + H3O+ * Acid has the extra proton (ability to donate it)
Assignment: Acid & Base Handout Questions 1 4
Buffer Solutions Buffers are solutions with the ability to resist the addition of strong acids or strong bases, within limits. eg. CH3COOH and its salt NaCH3COO Most buffer solutions are made up using a WEAK ACID and its sodium salt!
When a strong base such as NaOH is added to the buffer CH3COOH … • CH3COOH reacts with and consumes the excess OH- •OH- reacts with the H3O+ ion from the acid in the following reaction: H2O + CH3COOH <---> H3O+ + CH3COO- H3O+ + OH- <--------> H2O
When a strong acid such as HCl is added to the buffer CH3COOH … • H3O+ reacts with the CH3COOH- ions of the salt and form more undissociated CH3COOH H3O+ + CH3COO- <-------> H2O + CH3COOH There is a limit to the quantity of H+ or OH- that a buffer can absorb without undergoing a significant change in pH.
Buffer Components A buffer has two components. HANaA ---> Na+ + A- (weak acid) (a soluble salt of the weak acid) Therefore any extra H3O+ will be neutralized by the A- in the buffer H3O+ + A- <-------> HA + H2O And any extra OH- that is added will be neutralized by the acid HA + OH- <------> A- + H2O
Sample Problem: CH3COOH + H2O CH3COO+ H3O+ 1.0 M 1.0M Original Ratio: Acid = 1.0 = 1.0 Base 1.0 New Ratio: 1.1 = 1.22 0.9 Add 0.1 mol H3O+ ACID 1.1 BASE 0.9
Using Ka Original [H3O+] = (1.8 x 10-5)(1.0) = 1.8 x 10-5 pH = 4.74 After [H3O+] = (1.8 x 10-5)(1.22) = 2.5 x 10-5 pH = 4.66 pH change of 0.08
With no buffer present… Adding 0.1 mol of acid to H2O would change the pH by 6.00! pH water = 7 pH water with 0.1M H3O+ = 1 Thus a difference of 6.