# Buffer Solutions (OCR A2 Chem) =[

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#1
Hi,

please can someone help me with this exam question/ explain buffer solutions generally to me. I don't know what year its from, its just in a pack the school put together.

'Superphosphate' fertilisers contan calcium dihydrogenphosphate. Ca(H2PO4)2. This compound is one of the worlds most important ferilisers. When dissolved in water, Ca(H2PO4)2 dissociate forming H2PO4- ions which are easily taken up by plants.

a) Calcium dihydrogenphosphate, Ca(H2PO4)2 is produced by treating rock phsophate Ca3(PO4)2 with sulphuric acid H2SO4.

Write a balanced equation for this reaction.

Got:
Ca3(PO4)2 + 2H2SO4 - > Ca(H2PO4)2 + 2CaSO4

b) Aqeuous H2PO4- ions can act as a weak acid, write an equation to represent the dissociate of the H2PO4- ion

Got:
H2PO4- -> 2H+ + PO4- (reversible)

c) i) The H2PO4- ion can act as either an acid or a base.
State the formula of the conjugate base of H2PO4-

HPO4^2-

ii) State the formula of the conjugate acid of H2PO4-

H3PO4

iii) A solution of calcium dihydrogenphosphate Ca(H2PO4)2 in water acts as a buffer solution.
Suggest, with the aid of equations, how this buffering action takes place.

I have no idea what to write. I hate buffer solutions. Arrgh.

I would guess that, being in water, you could have water as a weak acid with this:

H2O -> H+ + OH-

but I have no idea how that really relates to anything
0
10 years ago
#2
(Original post by kej817)
Hi,

please can someone help me with this exam question/ explain buffer solutions generally to me. I don't know what year its from, its just in a pack the school put together.

'Superphosphate' fertilisers contan calcium dihydrogenphosphate. Ca(H2PO4)2. This compound is one of the worlds most important ferilisers. When dissolved in water, Ca(H2PO4)2 dissociate forming H2PO4- ions which are easily taken up by plants.

a) Calcium dihydrogenphosphate, Ca(H2PO4)2 is produced by treating rock phsophate Ca3(PO4)2 with sulphuric acid H2SO4.

Write a balanced equation for this reaction.

Got:
Ca3(PO4)2 + 2H2SO4 - > Ca(H2PO4)2 + 2CaSO4

b) Aqeuous H2PO4- ions can act as a weak acid, write an equation to represent the dissociate of the H2PO4- ion

Got:
H2PO4- -> 2H+ + PO4- (reversible)

c) i) The H2PO4- ion can act as either an acid or a base.
State the formula of the conjugate base of H2PO4-

HPO4^2-

ii) State the formula of the conjugate acid of H2PO4-

H3PO4

iii) A solution of calcium dihydrogenphosphate Ca(H2PO4)2 in water acts as a buffer solution.
Suggest, with the aid of equations, how this buffering action takes place.

I have no idea what to write. I hate buffer solutions. Arrgh.

I would guess that, being in water, you could have water as a weak acid with this:

H2O -> H+ + OH-

but I have no idea how that really relates to anything
First I would define what a buffer is, i.e mixtures that resists changes in ph on small additions of acid/alkali. Then show two/three equations:

H3PO4 -> H2PO4- + H+ should be equilibrium
HPO4 2- + H+ -> H2PO4 - also should be in eq

You can then use the H+ + OH- -> H20

I think you should also state what acts as a weak acid and what is the conjugate base.
0
10 years ago
#3
Does anyone have the OCR chemistry unit 5 jan 2011 mark scheme?
0
10 years ago
#4
This doesn't answer this particular question. But in general for a 7 mark question asking you to explain how a buffer system works. This method seems to work

1)Define a buffer system

2)Give the equation for the buffer equilibrium in the question

3)State what would happen to the equilibrium if an acid was added (as in [H+] increases, conjugate base reacts with H+ ions, the equilibrium shifts to the left)

4)Give the equation for this again if you want with an arrow showing the equilibrium shift.

5)State what would happen if an alkali was add. (i.e. [OH-] increases and a small concentration of H+ ions react with the OH- ions

6) Then give this equation: H+(aq) + OH- (aq) -->H20 (l)

7) State that the acid dissociates shifting equilibrium to the right to restore most of the H+ ions that reacted (again you have the option of rewriting out the equilibrium with an arrow to show the change but this isn't essential.)
1
10 years ago
#5
Also to work out the pH of a buffer solution:

Ka=[H+][A-]/[HA]

[HA] at equilibrium is taken to be the same as the original concentration as only a small quantity of it has dissociated.

[A-] is taken to be equal to the concentration of the salt of the weak acid.

You should also be given Ka.

Put the number in the equation to find [H+]

Then use pH=-log[H+] to find pH
0
5 years ago
#6
Can someone explain to me how you got to the answer for B please?
0
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