acids and bases
- Created by: Nataliagx
- Created on: 29-11-18 09:59
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- Acids & Bases
- Arrhenius Theory
- Acids: substances that produce hydrogen ions in solution.
- Bases: substances that produce hydroxide ions in solution.
- Neutralisation is from H+ ions and OH- ions that react to produce H2O
- However, when HCl reacts with NH3, no OH-ons are needed
- NH3 + H2O is a reversible reaction- that is how OH-ions are generated
- However, when HCl reacts with NH3, no OH-ons are needed
- Bronsted-Lowry Theory
- Acid: Proton (H+) donor
- Base: Proton (H+) acceptor
- ammonia = base; accepts proton (H+) that is attached to a lone pair on the nitrogen of NH3
- Strong acids:
- HCl
- H2O + HCl > H3O+ + Cl-
- FORWARD: HA (HCl) is an acid. H2O is a base
- REVERSE: H3O+ is an acid and A- (Cl-) is a base
- The reversible reaction contains 2 acids and 2 bases: CONJUGATE PAIRS
- conjugate pairs differ by the presence/ absence of the H+.
- H2O acts as acid & base: AMPHOTERIC
- H3O+ is a very strong acid,
- Kw= [H3O+] x [OH-]
- OH- is a very strong base
- little of the water is ionised and so, the conc remains unchanged
- varies with temperature
- 1.00 x 10-14 mol2 dm-6 at room temperature
- INCREASE temp = FORWARD reaction favoured, more H+ and OH- formed. This increases f Kw with temp.
- pH of pure water FALLS as the temp INCREASES
- although the pH changes, it still remains NEUTRAL
- 1.00 x 10-14 mol2 dm-6 at room temperature
- pKw = -log(Kw)
- pKw at room temp = 14
- 1. find [H+] or [OH-] (they're the same thing)
- at room temp, Kw is:[H+] [OH-] = 1.00 x 10-14
- but as [OH-] = [H+] it can be written as [H+]2 = 1.00 x 10-14
- then square root: [H+] = 1.00 x 10-7 mol dm-3
- pH = -log [H+]
- then square root: [H+] = 1.00 x 10-7 mol dm-3
- but as [OH-] = [H+] it can be written as [H+]2 = 1.00 x 10-14
- at room temp, Kw is:[H+] [OH-] = 1.00 x 10-14
- Kw= [H3O+] x [OH-]
- OH- is a very strong base
- H3O+ is a very strong acid,
- H2O + HCl > H3O+ + Cl-
- HNO3
- HBr
- H2SO4
- HCl
- Weak aids
- HSO4-
- HNO2
- C6H5COOH
- H3PO4
- pH
- concentration of hydrogen ions in a solution.
- pH = -log [H+]
- STRONG ACIDS: fully dissociate
- if the conc of acid is 0.1 mol dm-3, then the conc of H+ is also 0.1 mol dm-3.
- WEAK ACIDS: partially dissociates
- The further to the left Eq is, the weaker the acid is.
- Ka = [H+] [A-] / [HA]
- Weak acid= small Ka
- pKa= -log(Ka)
- low pKa = Strong acid
- Ka= 10 (pka)
- Ka = [H+] [A-] / [HA]
- Weak acid= small Ka
- pKa= -log(Ka)
- low pKa = Strong acid
- Ka= 10 (pka)
- Ka = [H+] [A-] / [HA]
- pH curve & Titration
- indicator changes colour= END POINT
- Weak acid: LITMUS PAPER
- OH- ions
- H+ ions
- OH- = H+
- Weak acid: LITMUS PAPER
- equation proportions = EQUIVALENCE POINTS
- indicator changes colour= END POINT
- STRONG ACIDS: fully dissociate
- pH = -log [H+]
- concentration of hydrogen ions in a solution.
- Buffer Solutions
- resists changes in pH when small amounts of acid or alkali is added
- Acid Buffer= pH lower than 7; weak acid + salt
- Ethanoic acid (weak acid); equilibrium shift left.
- ethanoate ions from the sodium ethanoate
- un-ionised ethanoic acid
- H+ combine w/ the ethanoate ions = ethanoic acid.
- ethanoate ions from the sodium ethanoate
- un-ionised ethanoic acid
- H+ combine w/ the ethanoate ions = ethanoic acid.
- H+ combine w/ the ethanoate ions = ethanoic acid.
- un-ionised ethanoic acid
- ethanoate ions from the sodium ethanoate
- H+ combine w/ the ethanoate ions = ethanoic acid.
- un-ionised ethanoic acid
- enough H+ to make the solution acidic
- Alkali removal w/ ethanoic
- Acid reacts w/ OH- to make ethanoate ions + H2O
- Alkali removal w/ H+
- OH- combine with H+ from ethanoic acid = H2O
- Eq shifts RIGHT to replace lost H+ ions.
- OH- combine with H+ from ethanoic acid = H2O
- ethanoate ions from the sodium ethanoate
- 0.10 mol dm-3 of ethanoic acid and 0.20 mol dm-3 of sodium ethanoate. pH?
- Ka = [H+] [A-] / [HA]
- assume that the ethanoate ion conc = conc of sodium ethanoate; 0.20 mol dm-3.
- [H+] = Ka x [acid]/[salt]
- Ka for ethanoic acid is 1.74 x 10-5 mol dm-3
- to calculate PROPORTIONS- reverse it
- 1. In-log the pH (10) = [H+]
- 2. Ka / [H+]
- conc of ethanoate ions (from the sodium ethanoate) in the solution has to be 0.5 times that of the conc of the acid.
- 2. Ka / [H+]
- 1. In-log the pH (10) = [H+]
- to calculate PROPORTIONS- reverse it
- Ka = [H+] [A-] / [HA]
- Ethanoic acid (weak acid); equilibrium shift left.
- Alkali Buffer= pH higher than 7; weak base + salt
- ammonia + ammonium chloride: weak base = Eq shift LEFT
- unreacted ammonia
- ammonium ions from the ammonium chloride
- enough OH- to make the solution alkaline
- Acidic Removal w/ ammonia
- NH3 + H+ = NH4 +
- Acidic removal w/ OH-
- H+ combine w/ OH- to make H2O
- Eq shifts RIGHT to restore OH-
- H+ combine w/ OH- to make H2O
- pH w/ 0.100 mol dm-3 of ammonia and 0.0500 mol dm-3 of ammonium chloride
- Ka ammonium ion is 5.62 x 10-10 mol dm-3
- [H+]= [alkali]/[salt]
- Ka ammonium ion is 5.62 x 10-10 mol dm-3
- ammonia + ammonium chloride: weak base = Eq shift LEFT
- Acid Buffer= pH lower than 7; weak acid + salt
- resists changes in pH when small amounts of acid or alkali is added
- Arrhenius Theory
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