Flashcards
21.1 Buffer Solutions
Flashcards do conjunto (29)
- 1What is a buffer solution?System that minimises the pH changes when small amounts of an acid or a base are added
- 2What are the two main components buffer solutions contain to remove added acid or alkali?A weak acid (as component 1) and its conjugate base (as component 2)
- 3How do you represent the weak acid and what does it do?HA
Removes added alkali - 4How do you represent the conjugate base and what does it do?A- removes added acid
- 5What happens when acid and alkalis are added to a buffer?The two components in the buffer solution react and will eventually be used up
As soon as one component has been reacted, the solution loses its buffering ability towards added acid or alkalis - 6As the buffer works does the pH change?pH does change but only by a small amount - pH does not stay completely constant
- 7Preparation from a weak acid and its salt - What can you mix together?A buffer solution can be prepared by mixing a solution of ethanoic acid, CH3COOH with a solution of one of its salts, for example, Sodium ethanoate CH3COONa
- 8Preparation from a weak acid and its salt - What happens when ethanoic acid is added to water? Write the equationThe acid partially dissociates and the amount of ethanoate ions in solution is very small
Ethanoic acid is the source of the weak acid component of the buffer solution
CH3COOH(aq) ⇌ H+(aq) + CH3COO-(aq)
component 1 - 9Preparation from a weak acid and its salt - What are salts of weak acids and what do they provide?Ionic compounds and provide a convenient source of the conjugate base
When added to water, the salt completely dissolves and so the salt is the source of the conjugate base component of the buffer solution - 10Preparation from a weak acid and its salt - Write the equation for the dissociation of salts of weak acidsCH3COONa(s) + aq → CH3COO-(aq) + Na+(aq)
component 2 - 11Preparation by partial neutralisation of the weak acid - What can you add to the weak acid instead?A buffer solution can also be prepared by adding an aqueous solution of an alkali e.g. NaOH(aq) to an excess of the weak acid
- 12Preparation by partial neutralisation of the weak acid - How does this work?The weak acid is partially neutralised by the alkali, forming the conjugate base
Some of the weak acid is left over unreacted - 13Preparation by partial neutralisation of the weak acid - What does the resulting solution contain?The resulting soluion contains a mixture of the salt of the weak acid and any unreacted weak acid
- 14Two reservoirs to remove added acid and alkali - In the ethanoic acid equilibrium where does the equilibrium position lie?Left - towards ethanoic acid
- 15Two reservoirs to remove added acid and alkali - What happens when you add CH3COO- ions to CH3COOH?The equilibrium position shifts further to the left, reducing the already small concentration of H+(aq) ions, and leaving a solution containing mainly two components, CH3COH and CH3COO-(aq)
- 16Two reservoirs to remove added acid and alkali - Write the equation of dissociation of CH3COOH(aq) and label the components 1 and 2.CH3COOH(aq) ⇌ H+(aq) + CH3COO-(aq)
component 1 component 2
CH3COOH and CH3COO- act as two reservoirs that are able to act independently to remove added acid and alkali
This is achieved by shifting the buffers equilibrium system either to the right or left - 17Action of the buffer solution - The conjugate acid-base pair, HA(aq) / A-(aq) in an acid buffer solution controls the pH Write the equilibrium and say how you can explain control of pH?HA(aq) ⇌ H+(aq) + A-(aq)
The control of pH can be explained in terms of shifts in the equilibrium position using le Chatelier's principle - 18Action of the buffer solution - Conjugate base removed added acid. Explain how.On addition of an acid, H+(aq):
[H+(aq)] increases
H+(aq) ions react with the conjugate base, A-(aq)
The equilibrium position shifts to the left, removing most of the H+(aq) ions
HA(aq) ⇌ H+(aq) + A-(aq)
<-------------------
added acid - 19Action of the buffer solution - Weak acid removed added alkali. Explain how.On addition of an alkali, OH-(aq):
[OH-(aq)] increases
The small concentration of H+(aq) ions reacts with the OH-(aq) ions:
H+(aq) + OH-(aq) → H2O(l)
HA dissociates, shifting the equilibrium position to the right to restore most of the H+(aq) ions
HA(aq) ⇌ H+(aq) + A-(aq)
------------->
added alkali - 20pH and buffer solutions - choosing the components for a buffer solution Different weak acids result in buffer solutions that operate over different pH ranges. When is buffer most effective?A buffer is most effective at removing either added acid or alkali when there are equal concentrations of the weak acid and its conjugate base
- 21pH and buffer solutions - choosing the components for a buffer solution Different weak acids result in buffer solutions that operate over different pH ranges. When [HA(aq)] = [A-(aq)] ..The pH of the buffer solution is the same as the pKa value of HA
The operating pH is typically over about two pH units, centred at the pH of the pKa value - 22Calculating the pH of a buffer solution - Write the equilibrium and Ka expression for a weak acid.HA(aq) ⇌ H+(aq) + A-(aq)
Ka = [H+(aq)] [A-(aq)]
____________________
[HA(aq)] - 23Calculating the pH of a buffer solution - What approximation do you make which doesn't work here?When you calculate the pH of a weak acid, you make an approximation that [H+(aq)] = [A-(aq)]
For a buffer solution, this is no longer true as A-(aq) has been added as one of the components of the buffer - 24Calculating the pH of a buffer solution - Rearrange the Ka expression to find [H+(aq)][H+(aq)] = Ka x [HA(aq)]
_________
[A-(aq)] - 25Calculating the pH of a buffer solution - What do you need to be able to calculate the pH?Provided that Ka and concentrations of HA and A- are known
[H+(aq)] and the pH can be calculated - 26Calculating the pH of a buffer solution - What about when the concentrations of HA and A- are the same?[HA(aq)] = [A-(aq)] and you cancel the A-(aq) on the top and HA(aq) on the bottom
Ka = [H+(aq) and pKa = pH - 27Calculate the pH when the buffer solution contains 0.100moldm-3 CH3COOH and 0.300moldm-3 CH3COONa. Ka(CH3COOH) = 1.74 x 10^-5 moldm-31) Calculate [H+(aq)] from Ka, [HA(aq)], and [A-(aq)]
[H+(aq)] = Ka x [CH3COOH(aq)]
__________________
[CH3COO-(aq)]
= 1.74 x 10^-5 x 0.100
--------
0.300
= 5.80 x 10^-6 moldm-3
2) Use your calculator to find pH
pH = -log[H+(aq)] = -log(5.80 x 10^-6) = 5.24 - 28150cm3 of 1.00moldm-3 HCOOH is mixed with 100cm3 0.750moldm-3 HCOONa. Calculate the pH of the buffer solution formed. Ka(HCOOH) = 1.78 x 10^-4moldm-31) Calculate the amounts in mol of HCOOH and HCOO- in the buffer solution
n(HCOOH) = (1.00 x 150) / 1000 = 0.150 mol
n(HCOO-) = (0.750 x 100) / 1000 = 0.0750 mol
2) Calculate the concentrations of HCOOH and HCOO- in the buffer solution
Total volume of buffer solution = 250cm3
[HCOOH(aq)] = (1000 x 0.150) / 250 = 0.600moldm-3
[HCOO-(aq)] = (1000 x 0.0750) /250 = 0.300moldm-3
3) Calculate [H+(aq)] from Ka, [HA(aq)] and [A-(aq)]
[H+(aq)] = Ka x [HCOOH(aq)] / [HCOO-(aq)]
= 1.78 x 10^-4 x (0.600 / 0.300)
= 3.56 x 10^-4 moldm-3
4) Use your calculator to find the pH
pH = -log[H+(aq)] = -log(3.56 x 10^-4) = 3.45 - 29100cm3 of 0.750moldm-3 NaOH(aq) is added to 150cm3 of 1.50moldm-3 HCOOH. Calculate the pH of the buffer solution formed. Ka(HCOOH) = 1.78 x 10^-4 moldm-31) Calculate the amount, in mol, of HCOO- in the buffer solution
The partial neutralisation is:
HCOOH(aq) + NaOH(aq) → HCOONa + H2O
n(NaOH) added = (0.750 x 100) / 1000 = 0.0750mol
n(HCOO-) formed = 0.0750mol
2) Calculate the amount, in mol of HCOOH in the buffer solution
n(HCOOH) used = (1.50 x 150) / 1000 = 0.225mol
n(HCOOH) remaining = n(HCOOH) used - n(NaOH) added
= 0.225 - 0.0750 = 0.150mol
and then do what you did in the last steps
Informações
- Descrição
- A level Chemistry (Module 5: Physical chemistry and transition elements) FlashCards sobre 21.1 Buffer Solutions, criado por Urvi B em 19-01-2019.
- Criado por
- Urvi B
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