Acid-base Equilibria Flashcards
Brønsted-Lowry acid
Proton donor
H+ combines with H2O to form H3O+
Brønsted-Lowry base
Proton acceptor
Forms water
Strong acid
Dissociate almost completely in water
Weak acid
Dissociate only slightly in water
Equilibrium is set up where equilibrium lies to the left
Conjugate pairs
Species that are linked by the transfer of a proton
Conjugate base
Species that has lost a proton
Conjugate acid
Species that has gained a proton
Neutral solutions
[H+]=[OH-]
All of each ion reacts to form water
Standard enthalpy change of neutralisation
Enthalpy change when solutions of an acid and a base react together under standard conditions to produce one mole of water
Define pH
Measure of the hydrogen ion concentration
Calculating pH
-log10[H+]
(Don’t type the 10 in calculator)
Calculating [H]+ from pH
10^-pH
Weak base
Partially react with water to form OH-
Calculating pH of a strong acid
As they fully dissociate, [acid]=[H+]
Calculating pH of a weak acid
Don’t fully dissociate so need Ka and [acid]
Assume that the [H+]=[X-]
Ka= [H+]^2 / [acid]
Monoprotic
Each mole of acid produces one mole of H+
Poly protic
Each mole of acid produces more than one mole of H+
Assumption from textbook for calculating pH of weak acids
Dissociation of acid is much stronger than that of water (all H+ come from the acid)
[acid at start]=[acid at equilibrium]
[base]=[H+]
Calculating pH of a strong base
Fully dissociate to release OH-
[OH-]=[base]
Need Kw and [OH-] to calculate [H+] then pH
Ionic product of water
H2O—-> H+ + OH-
<—-
So Kc expression applies
Water only dissociates a tiny amount so equilibrium lies to left so consider [H2O] to have a constant value. So Kc multiplied by [H2O] makes a constant, Kw
Pure water, [H+]={OH-] so [H+]^2
Define pKa and pKw
PKw=-log10Kw
Kw=10^-pKa
PKa=-log10Ka
Ka=10^-pKa
Calibrating a pH metre
Place the bulb of the pH meter into deionised water and allow the reading to settle and adjust the reading so that it says 7
Repeat for a standard solution of pH4 and 10 (rinsing the meter with deionised water between each reading)
PH of salts of strong acids and strong bases
Neutral
Using masses and pH to calculate Ka for weak acids
Calculate moles of the acid
Calculate concentration of the acid solution
Use pH to work out concentration of H+ at equilibrium
Ka=[H+]^2/[HA]
Remember [HA] start=[HA]equilibrium
Diluting a strong acid
Diluting by a factor of:
10: increases pH by 1
Diluting a weak acid by a factor of 10
Formula for pH of a weak acid
increases pH by 0.5
PH=-log10 root(Ka[acid])
Titration curves
Plot pH against amount of base added
Equivalence line has the equivalence point, where [H+]=[OH-], neutralisation. Small addition of base causes huge change in pH
Initial pH depends on strength of the acid
Final pH depends on strength of the base
Using titration curves to find pKa
Find the pH at half the equivalence point
Practical
Rinse burette with distilled water then with the solution filling it
Calibrate the pH probe and rinse it with distilled water
Perform rough titre and record pH in increments of volume of solution in burette to find approximate equivalence point
(PH when just pink if using indicator then add more of the acid and record that pH)
Repeat, decreasing size of increments as getting close to rough titre
Graph of pH against volume of burette solution added
Kw at 298K
1x10^-14 mol^2dm^-6
Amphoteric
Acts as a base and an acid
Water
