Acids and bases Flashcards
A Bronsted-Lowry acid
A proton donor, an acid dissociates and releases H+ into an aqueous soloution
A Bronsted-Lowry base
A proton acceptor, alkalis dissociate and release OH- ions into aqueous soloutions
Conjugate acid-base pair
A conjugate acid-base can be interconnected by the transfer of a proton, ie HCl or Cl-
Monobasic
Contains one hydrogen ion
Dibasic
Contains two hydrogen ions
Tribasic
Contains three hydrogen ions
Half equation of acid and metal
2H+ + Zn –> Zn+2 + H2
Calculation changes in pH of the buffer
If a small amount of alkali is added to a buffer then the moles of the buffer acid would reduce by the number of moles of alkali added and the moles of salt would increase by the same amount so a new calculation of pH can be done with the new values. If a small amount of acid is added to a buffer then the moles of the buffer salt would reduce by the number of moles of acid added and the moles of buffer acid would increase by the same amount so a new calculation of pH can be done with the new values.
Half equation of acid and carbonate
2H+ + CO3-2 –> H2O + CO2
Half equation of acid with metal oxide
2H+ + MgO –> Mg+2 + H2O
Half equation of acid and alkali
H+ + OH- –> H2O
pH of strong acid
You assume it fully dissociates so if it is a monobasic acid its concentration will equal the concentration of H+
Working out pH
-log10[H+]
Working out the concentration of H+
10^-pH
Ka
[HA]
pKa
-log10[ka]
Approximations about ka
That HA dissociates to produce an equal concentration of A- and H+ and the H+ from the dissociation of water will be neglected because it is so small. The second assumption is that because the dissociation of HA is so small it can be discounted, so you assume that the concentration of HA is the same at the start as it is at equilibrium
What’s the wrong about the first ka approximation
When the pH is very weak i.e. above 6, then the dissociation of water will be significant when compared to the weak acid
What’s wrong with the second ka approximation
The dissociation of the weak acid will be significant for stronger weak acids as there will be a real difference between the concentration of HA at the start and at equilibrium
The ionic product of water
Kw = [H+] x [OH-]
Value of Kw
1 x10^-14
This value is constant at 25 degrees
Figuring out the pH of a strong base
For bases we are normally given the concentration of the hydroxide ion, as we assume the base completely dissociates to give hydroxide ions. To work out the pH we need to work out [H+(aq)] by rearranging the kw expression
Equation of simplified Ka
[H+ (aq)]2
——————- = Ka
[HA (aq)]initial
To find out the pH of a diluted strong acid
[H+] = [H+]old x old volume
—————————-
new volume
pH = – log [H+]