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# Buffer solutions question watch

1. A buffer solution was prepared by mixing 0.1 moldm3 ethanoic acid with sufficient sodium ethanoate so that the final concentration of sodium ethanoate was also 0.1 moldm3. The total vol. of the buffer was 100cm3. pka of ethanoic acid is 4.76.

Calculate:
-Mass of sodium ethanoate needed to make the buffer
-ph of the buffer

2. (Original post by cocoabeaneo)
A buffer solution was prepared by mixing 0.1 moldm3 ethanoic acid with sufficient sodium ethanoate so that the final concentration of sodium ethanoate was also 0.1 moldm3. The total vol. of the buffer was 100cm3. pka of ethanoic acid is 4.76.

Calculate:
-Mass of sodium ethanoate needed to make the buffer
-ph of the buffer

mass is simply the mass required to make 100ml of a 0.1 mol dm-3 solution.

When acid concentration = salt concentration pKa = pH
3. (Original post by charco)
mass is simply the mass required to make 100ml of a 0.1 mol dm-3 solution.

When acid concentration = salt concentration pKa = pH
But how would you work out the ph of the buffer? The ph given is just for the acid
4. (Original post by cocoabeaneo)
But how would you work out the ph of the buffer? The ph given is just for the acid
I repeat:

When the concentration of acid and salt are the same, the pKa of the acid = pH of the solution.
5. (Original post by charco)
I repeat:

When the concentration of acid and salt are the same, the pKa of the acid = pH of the solution.

One more question: if the solution is diluted will the pH be the same again?
6. (Original post by cocoabeaneo)

One more question: if the solution is diluted will the pH be the same again?
yes
7. Another Buffer soln question here
Cacuate the p(OH) and the pH of an aqueous ammonia given that kb+1.8.10^-3 at 298k
Another Buffer soln question here
Cacuate the p(OH) and the pH of an aqueous ammonia given that kb+1.8.10^-3 at 298k
9. (Original post by charco)
kb=[NH4][OH-]/[NH4OH]
then
-log[OH-]=-logkb+log[NH4+]/[NH4OH]
since -log[OH-]=POH,POH+PH=14,then
the equation will be
14-PH=-log[1.8 times 10^-3]+log[NH4+]/0.1
and that is where I got stuck

Calcuate the p(OH) and the pH of an aqueous ammonia given that kb+1.8x10^-3 at 298k this should read 1.8 x 10[sup-5[/sup]

kb=[NH4][OH-]/[NH4OH]

then
-log[OH-]=-logkb+log[NH4+]/[NH4OH] this line is confusing you. There is no need to go into the log form yet.

since -log[OH-]=POH,POH+PH=14,then
the equation will be
14-PH=-log[1.8 times 10^-3]+log[NH4+]/0.1
and that is where I got stuck
kb=[NH4+][OH-]/[NH4OH]

and you know that according to the stoichiometry of the reaction:

NH3 + H2O --> NH4+ + OH-

That [NH4+] = [OH-]

hence

kb * [NH4OH] =[OH-]2

assuming that the ammonia solution is 0.1 mol dm-3 (you forgot to state it)

and that very little interacts with the water ( a good approximation)

then

1.8 x 10-5 * 0.1 = [OH-]2

[OH-] = √(1.8 x 10-6)

[OH-] = 1.34 x 10-3

pOH = 2.87

pH = 11.13
11. (Original post by charco)
kb=[NH4+][OH-]/[NH4OH]

and you know that according to the stoichiometry of the reaction:

NH3 + H2O --> NH4+ + OH-

That [NH4+] = [OH-]

hence

kb * [NH4OH] =[OH-]2

assuming that the ammonia solution is 0.1 mol dm-3 (you forgot to state it)

and that very little interacts with the water ( a good approximation)

then

1.8 x 10-5 * 0.1 = [OH-]2

[OH-] = √(1.8 x 10-6)

[OH-] = 1.34 x 10-3

pOH = 2.87

pH = 11.13
Thank you so much
Thank you so much
And sorry to bother again but when do u use the stoichiometry concept?and how do u use it?
And sorry to bother again but when do u use the stoichiometry concept?and how do u use it?
In any dilute weak acid or base situation.

You can't use it with buffers.
14. (Original post by charco)
In any dilute weak acid or base situation.

You can't use it with buffers.
ohkay gat it now thank you

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