- Created by: spooky0114
- Created on: 24-04-14 15:47
ACID: Proton donor
BASE: Proton Acceptor
Acid-Base equilibria involves the transfer of protons.
Dissociation in water STRONG ACIDS (HCl): Completely dissociate in water STRONG BASES (NaOH): Ionise completely in water WEAK ACIDS (CH3COOH): Partially dissociate in water WEAK BASES (NH3): Partially dissociate in water
Ionic Product of Water
Kw = [H+][OH-] H2O H+ + OH-
Kc = [H+] [OH-] . [H2O] Water only dissociates a small amount Kw = 1.00 x 10^-14
pH = -Log10[H+] [ ] ---> concentration in mol dm-3
[H+] from pH If you have pH and want to know [H+] [H+] = 10 ^-pH
pH of strong monoprotic acids Monoprotic --> each molecule of an acid releases on proton when it dissociates. HCl ---> H+ + Cl- e.g. calculate pH of 0.100 moldm-3 HCl : -Log(0.0100) = 1.00
pH of a strong diprotic acid Diprotic--> each molecule of an acid releases 2 protons when it dissociates. H2SO4 ---> 2H+ + SO4^^2- e.g. calculate pH of 1.00 mol dm-3 H2SO4 0.100 x 2 -Log(0.200)=0.7
pH of strong bases NaOH and KOH ---> donates 1 OH- per mole of base Kw=[H+][OH-] 1.Find the value of Kw and [OH-], you may be told. 2. Rearrange the equation ---> [H+] == Kw / [OH-] 3. Know [H+] so do -Log[H+] to find pH
Acid Dissociation Constant
HA H+ + A- WEAK ACIDS
Ka = [H+] [A-] [HA]
Assume that [H+] is squared
pH: -Write Ka -Rearrange to find [H+]2 -take the square root of the value for [H+]2 then -Log[H+]
Concentration: -Find [H+] by 10^-pH -Write the Ka expression and then rearrange to find the concentration of the acid ---> [H+]2 / Ka -Substitute in the values and solve
pKa pKa = -Log10[Ka] Ka = 10^-pKa
Titrations and pH Curves
STRONG ACID + STRONG BASE:
STRONG ACID + WEAK BASE:
WEAK ACID + STRONG BASE:
WEAK ACID + WEAK BASE:
INDICATORS: Pick one changes colour over narrow pH. Methyl-Orange and Phenolphthalein
MONOPROTIC ACIDS: -Write out balancd equation -Decide what yyou know & what you need to know. -One reagent you know the conc. and vol. ---> calculate the moles of this : Moles = conc. (moldm-3) x ( Col. (cm3) / 1000) ) --Use molar ratios in balanced equations to find how many moes of the other reagent reacted. -Calculate the know conc. using: conc. = (mol x 1000) / Vol. (cm3)
DIPROTIC ACIDS: e.g. ethandioic acid. 2 stages to a pH curve with 2 equivalance points
You calculate the same way as monoprotic, BUT you need 2x's as many moles of base as moles of acid.
BUFFER: Resists changes in pH when small amounts of acid or alkali are added.
ACIDIC BUFFER pH < 7 Made by mixing weak acid with one of its salts.
Acid Added; -[H+] Increases -The rise in [H+] should decrease the pH but... Excess H= combines with A- and the equilibrium shifts left. -[H+] decreases to close to its original value --> pH same. Base Added; -[OH] increases. - OH- combines with H+ so [H+] decreases. -the equilibrium shifts right to replace the lost H+ -[H+] increases to close to its original value. --> pH stays almost same.
BASIC BUFFER pH > 7 Made by mixing weak base with one of its salts
Acid Added; -[H+] Increases -H+ combines with OH- so [OH-] decreases -Equilibrium shifts right to replace lost OH-. -[OH-] increases to close to its original value. --> pH stays almost the same. Base Added; -[OH-] increases -Excess OH- combines with BH+ and equilibrium shifts left. -[OH-] decreases to close to its original value.
APPLICATIONS: Shampoo, Biological Washing Powder and Blood
Calculation pH of Buffers
1. Write out Ka of Weak Acid 2. Rearrange to give expression for [H+] 3. Sub in the values. 4. Solve to find [H+] 5. -Log[H+]
1. Write out equation for neutralisation reaction Acid + Base ---> Salt + Water 2.Calculate moles of acid + base at start using Vol. and Conc. Mol = Conc. x (Vol. / 1000) 3.Use molar ratios to work which is in excess 4. Calculate Conc. of acid and salt in buffer solution by (Mols x 1000) / Vol. -Work out pH