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Acid-Base Equilibria

Acid- Base Equilibria


 

Note a low Ka value indicates really undissociated 

At half equivalence Ka = [H+]

 

What are the Acid Dissociation Constant (Ka) & the Base Dissociation Constant (Kb)

 

  • Used for WEAK ACID OR WEAK BASE respectively and this is because strong acids will completely ionise and so not be in equilibrium 

  • Eq for the equilibrium of a weak acid  is Ka = [H+][A-]/[HA]

  • Eq for the equilibrium Kb = [YH+][OH-]/[Y]  

  • Same method as writing standard equilibrium constants when the conc products multiplied together/conc reactants

  • Remember when doing calculation to root the value after the rearranging as we will likely have to assume that [H+] = [A-]

 

What is the Ionic Product of Water, Kw

 

  • Water is amphoteric and can act as a base or an acid 

  • Water can even self protonate (to form H3O+) this is called autoionisation 

  • The equilibrium constant for the self protonation is Kw 

  • Kw = [H+][OH-]

  • [H2O] is not included as we might expect as a result of the fact that the value is constant 

  • Value at 25˚C is 1 x 10^-14

  • At this temperature we can thus find the concentration of either [H+] or [OH-] as long as we are provided with the other 

 

NOTE: 

  • pH = -log[H+]

  • [H+] = 10^-ph

  • SO: definition of pH s the pH is equal to the negative logarithm to the base 10 of the hydrogen ion concentration 

 

What are pKw, pKb, pKa

 

  • Just remember that p means and represents the negative log of so to convert between pKw remember that pKw = -log Kw (applicable for all) 

  • Also note that 

    • pH + pOH = 14 at 25˚C 

    • pOH = 14 - pH at 25˚C

  • pKx = -log Kx 

  • Kx = 10^-pKx 

  • (x can be acid, base etc.) 

 

How do you calculate the pH of a Strong Base

 

  • Use the above and pOH = -logOH 

  • Using the calculated pOH value use pH = 14-pOH to get the value of the final pH

  • Alternatively use Kw and the known value of 1 x10^-14 

  • [H+] = 1 x10^-14 / [OH]

  • And then use -log[H+]

 

How do you calculate the pH of a weak acid:

  • Use standard ICE method to workout moles at start and moles at equilibrium 

  • Ka can be used and [H+] = [OH-]

  • Ka must be provided and then Ka = [H+]^2 / [HA]

  • NOTE: weak acids < 1

  • Assumptions made: 

    • [H+]=[A-]

    • [HA] = conc of unionised acid

    • Negligible [H+] from any H2O  

  • Ask for help with the explaining the logarithms part

 

How do you calculate the pH of a weak base:

 

  • Use Kb to calculate the [OH-] (same as with [H=] you will have to square root it)

  • pOH = -log[OH]

  • pH = 14 - pOH 

 

How do salts of weak acids affect the reaction/calculation

 

  • pH of a pure acid is calculated from Ka 

  • When a base/alkali is run in in a titration some of the acid will react to form a salt. The presence of this salt immediately means that [H+] cannot be equal to [A-] (salt) so the same method of calculating pH is immediately not applicable as the underlying assumption is incorrect 

  • Thus the method changes: 

    • Ka = [H+][A-]/[HA] can be rearranged to give [H+] = Ka [HA]/[A-]

    • Work out total volume of solution (acid vol + base vol given in question stem0

    • If equal conc of acid and base the difference between the vols is the amount of neutralised acid/base (say the total vol = 45 as 20cm3 base 25 cm acid of 0.1 mol dm-3 then 5 cm3 acid is unneutralised. 

    • Work out the conc of the remaining acid that is unneutralised by using the unneutralised volume divided by the total volume to gain a multiplication factor and then multiply this value by the concentration 

    • Use the same method to work out the new concentration of the salt ions (such as (ethanoate) which would be the same calculation using the volume of neutralised acid as the vol

    • Set up a ratio using the the calculated values ratio = [unneutralised HA]/[Salt]

    • Substitute this ratio in and multiply by the Ka to obtain a value for the [H+] 

    • -log[H+] 


 

Graphs:

 

  • Marked change in pH at equivalence point and it is this point that is key for detecting the end point and the indicator to use. The equivalence point is the point at which the amount in moles of the acid/base is exactly enough to react with the moles of the base/acid in the titration

  • The graphs overview:

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  • Strong acid and strong base: the pH will be equal to 7 as the base and acid fully dissociate forming water and salt 

  • Weak acid and strong base: PH will be above 7 at equivalence point 

  • In a titration of a strong acid and a weak base the pH at the equivalence point will be below 7 and after this point will rise to be less than with a strong acid and strong base 

  • With a weak acid and weak base the exact pH and equivalence points are hard to know but if the dissociation constants are approximately equal for the acid and base used then the pH will be close to 7 and neutral as the degree of dissociation will be similar.

  • The indicator used is vital and is based on equivalence:

    • Methyl Orange (Colour change at pH 3.2-4.4) Red-Yellow colour change

    • Methyl Red (Colour change at pH 4.2-6.3) Yellow-Red

    • Bromothymol blue (colour change at 6.0-7.6) Yellow-Blue 

    • Phenolphthalein (colour change at 8.2-10.0) Colourless-Pink 

  • Very little indicator is added so it does not affect the titration 

  • The indicator either gains or loses a proton to cause a colour change as it changes pH 

 

Buffer Solutions: 

 

  • Mixture of molecules and ions which help to keep the pH more or less constant when small quantities of an acid or alkali are added

  • Important i the food industry, naturally found in the body and environment 

  • Important as the pH will vary and this can affect the solution and stability or in food might make it go off sooner and more 

  • Also key in the pH of blood (bio for dissociation etc.)

  • Buffers are in equilibrium 

  • Solution will typically consist of a weak acid and one of its salts

  • Assures if more alkali added equilibrium shifts and acid dissociates more to form H+ ions to react with the lakli and take it out of solution. 

  • Assures if acid added enough salt ions to shift equilibrium to the left and react with the salt to maintain pH again 

  • If we choose a weak acid and salt of equal strength then Ka is valid and so the pH of the buffer can be calculated using the Ka 

  • We assume

    • ALL HA has come from the acid 

    • ALL A- has come from the salt 

  • Therefore 

    • [H+] = Ka x [Acid]/[Salt]

    • The affect of the buffer can often be seen on the graph of a titration as the pH will stay fairly constant until approaching the end point 

  • To calculate:

    • Ka can be looked up in data table 

    • Remember to recalculate the concentration if the question states that the salt has been added by doing the volume of the acid/salt over the total volume and multiplying this by the value of the concentration 

    • State assumptions 

    • Use [H+] = ka [acid]/[salt]

    • pH = -log [H+]

  • 2 Ways to make a Buffer: 

    • Titration (volume changes) (Ka = [Salt][H+]/[Acid]

    • Into a beaker of acid the solid salt is added (volume does not change) - slightly more control 

  • If they ask you to calculate the pH of a buffer after a small amount of acid/alkali was added in then remember that the equivalent moles of acid/salt is used so this decreases and the (other of the 2) salt/acid moles increases slightly and additionally the volume would have changed slightly so all these volumes would have been altered and so must be added in to get a final value for [H+] and thus pH 

  • If there is a lot of acid or a lot of base added workout the excess moles of the acid or base as the neutralisation will be 1:1 after this there is an excess. With this excess work out the new conc with this moles/the volume of acid and base added and then use this concentration and use either Ka equation or just -log[H+] if acid based or/then use pOH = -log[OH] and pH = 14 - pOH to get the final pH  

 

Neutralisation: 

  • Any reaction when an acid and base react to form a salt but pH does not have to be 7  

  • Neutralisation is essentially a reaction between aqueous hydrogen and hydroxide ions to form water 

  • If a strong base/acid and a weak base/acid are used then the pH cannot be 7 as the H will be biassed toward the stronger of the 2 

  • Standard enthalpy of neutralisation: (key here) enthalpy change when the acid and alkali in the equation reaction neutralise each other under standard conditions to form 1 mole of water 

  • Strong Acid & Bases:

    • Standard enthalpy of neutralisation for dilute solutions of strong acid and strong base is always close to -57.5 kJ mol-1 and the value is relatively constant because they are fully ionised/dissociated 

    • As a result the reaction is almost the same in that H+ and OH- react to produce water/H2O

    • Measured approximately in a calorimeter

  • Weak Acid & Bases:

    • Less negative standard enthalpy of neutralisation 

    • As not neutral at equivalence and not fully ionised so not the only reaction 

    • Could get asked a Q=mcdeltaT Q to calculate delta H etc. 

    • Typically around -55 kJ mol-1 

  • Reason for the values benign fairly similar is that the reaction occurs such that the H+ and OH- react and are removed form solution, this shifts equilibrium such that almost all of the weak acid reacts anyway to neutralise over time. 

  • Further explanation is that the enthalpy for the reaction of the H2O with the acid is very low in enthalpy and is almost 0 so barely affects the enthalpy 

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