Buffer Question

[rak173]

What is the pH of a buffer that is prepared by mixing 35 mL of 0.250 M acetic acid and 25 mL of 0.180 M NaOH? (pka of AA=4.74)

Do you use the HH equation even though I think you can only use weak acid and its conj. base in the equation. The answer is 4.76. Thanks.

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[hye340]

you sure that's the right answer

 

[rak173]

I know the answer but don't know to get to it. Can you please solve the problem step by step?

 

[m91]

You're right, you do use the HH equation since we recognize that it is a buffer solution.

Calculate the moles in each reactant
AA (.250 mol/L)(.035L)= 8.75 x 10^-3 mol AA
NAOH (.180mol/L)(.025L)=4.5x10^-3mol NaOH

As you can see, we have more AA and less NaOH, so NaOH is the limiting reactant.

Next, step up your ICE​

NaOH + CH3COOH -> NaCH3COOH + H2O​

start: 4.5 mmol, 8.75mmol, 0, n/a
change: -4.5 mmol, -4.5mmol, 4.5mmol, n/a
after rxn: 0, 4.25 mmol, 4.5mmol, n/a​

Now your HH equation

pKa + log(conjugate base/acid) = 4.74 + log(4.5x10^-3 / 4.25x10^-3) = pH = 4.76

(consulted my dusted chemistry book)

 

[flossit]

You're right, you do use the HH equation since we recognize that it is a buffer solution.

Calculate the moles in each reactant
AA (.250 mol/L)(.035L)= 8.75 x 10^-3 mol AA
NAOH (.180mol/L)(.025L)=4.5x10^-3mol NaOH

As you can see, we have more AA and less NaOH, so NaOH is the limiting reactant.

Next, step up your ICE​

NaOH + CH3COOH -> NaCH3COOH + H2O​

start: 4.75 mmol, 8.75mmol, 0, n/a
change: -4.75 mmol, -4.5mmol, 4.5mmol, n/a
after rxn: 0, 4.25 mmol, 4.5mmol, n/a​

Now your HH equation

pKa + log(conjugate base/acid) = 4.74 + log(4.5x10^-3 / 4.25x10^-3) = pH = 4.76

(consulted my dusted chemistry book)

Where did you get "start: 4.75 mmol" from. Please explain!

 

[m91]

Where did you get "start: 4.75 mmol" from. Please explain!

Oh I'm sorry, that was a typo. I fixed it now.

 

[tyjacobs]

What is the pH of a buffer that is prepared by mixing 35 mL of 0.250 M acetic acid and 25 mL of 0.180 M NaOH? (pka of AA=4.74)

Do you use the HH equation even though I think you can only use weak acid and its conj. base in the equation. The answer is 4.76. Thanks.

You could use the HH equation. Remember, you can only use the HH equation when you are in the buffer region (aka when [A-] and [HA] are within a factor of 10 of each other).

Another way of solving the problem is by using a PUG/ICE box. m91's did a good job of explaining how to solve this problem using this method in his post.

If you are confused, feel free to personal message me

 

[rak173]

You could use the HH equation. Remember, you can only use the HH equation when you are in the buffer region (aka when [A-] and [HA] are within a factor of 10 of each other).

Another way of solving the problem is by using a PUG/ICE box. m91's did a good job of explaining how to solve this problem using this method in his post.

If you are confused, feel free to personal message me

Hi, I never remember Chad mentioning that [A-] and [HA] should be within factor of 10 of each other. Is this true only for weak acid and its conj. base? The only thing Chad mentioned that to be a good buffer you should have 1:1 ratio of weak acid and its base or 2:1 of weak acid and strong base or 2:1 ratio of weak base and strong acid. Where does the "[A-] and [HA] are within a factor of 10 of each other" comes into play with that?

Thanks for the help.

 

[gabtse]

The HH equation can certainly be used regardless of what ratio of ionic species to molecular species you have. But to qualify as a good buffer, the ratio should stay within 1/10 or 10/1 so that:

pH = pKa plus or minus 1

since log (1/10) = -1 and log (10/1) = 1.

Here's the alternative method to arrive at the pH of the buffer:

pKa = -logKa
Ka = 10^-4.74 = 1.82 x 10^-5

[H+][A-]/[HA] = Ka
[H+] = [HA]Ka/[A-]
= (4.25)(1.82 x 10^-5/4.5)
= 1.72 x 10^-5

pH = -log 1.72 x 10^-5
= 4.76

 

[tyjacobs]

Hi, I never remember Chad mentioning that [A-] and [HA] should be within factor of 10 of each other. Is this true only for weak acid and its conj. base? The only thing Chad mentioned that to be a good buffer you should have 1:1 ratio of weak acid and its base or 2:1 of weak acid and strong base or 2:1 ratio of weak base and strong acid. Where does the "[A-] and [HA] are within a factor of 10 of each other" comes into play with that?

Thanks for the help.

On the DAT you won't have a calculator for the natural sciences section. So to calculate the ratio of the acid and its conjugate base or the base and its conjugate acid (depending on the problem given) will be quite annoying to do. So I imagine chad just gave a small mental shortcut you can use to help streamline the problem solving.

The HH equation, as you probably know, has 2 different forms:

It is important to note that the HH equation only applies when the pH or pOH is in the buffering region. The buffering region is the only region where the buffer exists. A buffer only exists when the acid and its conjugate base OR the base and its conjugate acid are within a factor of 10 of each other. Another way to look at it it is that the buffering region is equal to the pKa of the weak acid or pKb of the weak base +/- 1 pH unit. Conceptually, if we look at the following titration curve below (it assumes weak acid titrated with strong base), then region between A and C is the buffer region.

If the pH lies beyond the buffering region, then you cannot use the HH equation because there is no buffer present in the solution

Refer to this link: http://chemwiki.ucdavis.edu/Core/Ph...s/Buffers/Henderson-Hasselbalch_Approximation

It says the assumption that log ([A-] / [HA]) must be between -1 and 1 which also means [A-] and [HA] must be within a factor of 10 of one another



if you are confused, look at my gen chem natural science notes I go in depth on titrations