Acidic Buffer Calculations (A2 Chem) Help!! Watch

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Asphyxiating
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Could someone help me with this Chemistry question for AQA Module 4 (acidic buffer calculations).. I have got an answer but I don't think it's right and my exam is next week! Can someone just confirm that the answer is right because I don't think I've done it right Thanks!!

(indice)

*Find the ph if 3.28g of sodium ethanoate is added to 1dm(3) of 0.01M ethanoic acid, Ka = 1.84 x 10(-5)

this comes out to be ph= 5.34 (correct)

*now find the new pH after adding 10cm(3) 0.1M HCl.
I get ph = 6.32 for this, can someone confirm this because I don't think this answer is right..

thanks!
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hornblower
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Ah, Buffy solutions! I need to revise these as well...

i) Find the pH if 3.28 g of sodium ethanoate is added to 1 dm^3 of 0.01 M ethanoic acid, Ka = 1.84 * 10^(-5)

No. moles sodium ethanoate = mass / RFM = 3.28 / 82 = 0.04 mol

Now, Ka = [H+][salt] / [acid]

So, 1.84 * 10^(-5) = [H+] * 0.04 / 0.01

Thus, [H+] = 1.84 * 10^(-5) * 0.01 / 0.04 = 4.6 * 10^(-6) mol dm^(-3)

Therefore, pH = -lg [4.6 * 10^(-6)] = 5.34 (3 s.f.)

Tick and a smiley face


ii) Now find the new pH after adding 10 cm^3 of 0.1 M HCl.

No. moles H+ added = 0.1 * 10 / 1000 = 0.001 mol

Now, 0.001 mol of ethanoate ions is used up:

So, No. moles CH3COO- = 0.04 - 0.001 = 0.039 mol

But now, 0.001 mol of ethanoic acid is formed:

So, No. moles CH3COOH = 0.01 + 0.001 = 0.011 mol

Assume volume remains constant, i.e. 1 dm^3

Now, Ka = [H+][salt] / [acid]

So, 1.84 * 10^(-5) = [H+] * 0.039 / 0.011

Therefore, [H+] = 1.84 * 10^(-5) * 0.011 / 0.039 = 4.717... * 10^(-6) mol dm^(-3)

Hence, pH = -lg [4.717... * 10^(-6)] = 5.33 (3 s.f.)



The new pH is almost the same as the original - the Buffy solution does exactly what it says on the tin! And btw, isn't the Ka of ethanoic acid 1.74 * 10^(-5) and not 1.84 * 10^(-5) as you have quoted?

J.
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Asphyxiating
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Thanks, that's really a great help! I found out where I went wrong, with the units but somehow I get

[H+] = 1.84 * 10^(-5) * 0.011 / 0.039 = 5.1897..x 10(-6)
..instead of 4.717... * 10^(-6)?

I've got Ka as 1.84 x 10(-5) but hey maybe our teacher got it wrong

Anyway thanks! You make it look so easy.. now I know I am definetely going to fail this exam! :eek:
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Jonatan
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(Original post by Asphyxiating)
Thanks, that's really a great help! I found out where I went wrong, with the units but somehow I get

[H+] = 1.84 * 10^(-5) * 0.011 / 0.039 = 5.1897..x 10(-6)
..instead of 4.717... * 10^(-6)?

I've got Ka as 1.84 x 10(-5) but hey maybe our teacher got it wrong

Anyway thanks! You make it look so easy.. now I know I am definetely going to fail this exam! :eek:
It is actually quite simple if you know some equilibrium theory. Il take you through it step by step. First of all, a weak acid dissolved in water discociate to a very small extent. You get an equilibrium in the solution. If HA is an acid, the the general reaction looks like this:

HA <---> (H+) + (A-)

Now, if we considder the equilibrium of this reaction we have an equilibrium constant which, accourding to the definition of th equilibrium constant is:

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

This is simply the formula for the equilibrium constant for the particular reaction.

Suppose now we add a soluable salt, BA, to the solution. The salt discociate almost completely accourding to the equation

BA ---> (B+) + (A-)

Now, since the salt discociate completely, and because the weak acid discociate to such a low extent, almost all the [A-] ions will come from the salt. Therefore the aproximation [A-] = [salt] is a quite an accurate aproximation.

Also, since so few molecules of teh weak acid discociate, we may make another aproximation, namely that [HA] = [acid].

We do in other words have three different equations:
a) Ka = [H+][A-] / [HA]
b) [A-] = [salt]
c) [HA] = [acid].

Substituting b and c into a gives the familiar buffer solution formula:

Ka = [H+][salt]/[acid]

By rearrangeing this equation one obtains:

Ka [acid] / [salt] = [H+]

As you can see, the only chemistry you nedd to knwo is the expression for an equilibrium reaction and that a weak acid only discociate to a small extent. The rest is pure maths.
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