bolbol
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How to make a buffer of a specific pH?
This is a homework related problem that I am stuck on. Instructor is asking for a 50 mL solution that is a pH 9.35 buffer. The question is asking
"Calculate the volumes of 0.70 M NH3 and 1.0 M NH4Cl needed to prepare 50. mL of buffer with the pH of 9.35"
I did the math and I found the ratio of NH3/NH4+ required for that solution but I am not sure how to go on further from here. Have I approached it the
wrong way? If not how do I solve this problem?
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aga
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pH means the negative log of the concentration of H3O<sup>+</sup> in a solution.
-log(x) = 9.35, so 10<sup>-9.35</sup> is the concentration of H3O<sup>+</sup> required, which is hardly any.
pH - pOH = 14, so really what you're looking for is a pOH of 14 - 9.35 = 4.65.
pOH = -log(conc of OH<sup>-</sup> = 4.65
so the conc of OH<sup>-</sup> = 10 <sup>-4.65</sup>
Now you have the molar concentration of OH- required, so just have to calculate how much ammonia stuff to add to water to get that concentration.
(if i got that wrong i think i'll get whipped, again).
Edit:
Oh crap. Always read the question at least 5 times when drunk ....
NH3 is NH4<sup>+</sup> in aqueous solution.
Oh dear, i'm lost now. Let the whipping begin.
[Edited on 9-10-2016 by aga]
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DistractionGrating
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While it's not an answer to your question per se, the information in this link might help you confirm the validity of your calculations: http://delloyd.50megs.com/moreinfo/buffers2.html
EDIT: Sorry, never mind. Ammonia based buffers not covered in that link.
[Edited on 10/9/2016 by DistractionGrating]
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Darkstar
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You wouldn't happen to know anything about those two liters of absolute ethanol that mysteriously disappeared from B&D University's supply room
the other day, would you?
Quote: Originally posted by bolbol  | This is a homework related problem that I am stuck on. Instructor is asking for a 50 mL solution that is a pH 9.35 buffer. The question is asking
"Calculate the volumes of 0.70 M NH3 and 1.0 M NH4Cl needed to prepare 50. mL of buffer with the pH of 9.35"
I did the math and I found the ratio of NH3/NH4+ required for that solution but I am not sure how to go on further from here. Have I approached it the
wrong way? If not how do I solve this problem? |
It's been a while since I've actually done one of these problems, but I'm pretty sure the answer is 32.1 mL of NH3 and 17.9 mL of
NH4Cl. Here's how I got that answer:
Target buffer solution: 0.05 liters of 0.70 M NH3 and 1.0 M NH4Cl with a pH of 9.35.
To find the volumes needed, first calculate the actual concentrations of NH3 and NH4Cl in the final buffer solution.
Since we know that:
$$\mathrm{concentration}=\frac{\mathrm{moles}}{\mathrm{volume}}$$
And:
$$\mathrm{moles
}=\mathrm{volume
}\times \mathrm{concentration}$$
We can find the concentration of NH3 in the buffer solution by multiplying the concentration of the ammonia solution by its initial volume
and then dividing by the new volume of the buffer solution. Let x represent the unknown volume of the ammonia solution:
$$[NH_3]=\frac{x(0.70~\mathrm{M})}{0.05~\mathrm{L}}$$
Similarly, we can also find the concentration of NH4Cl in the buffer solution this way as well. Since x is the volume of the ammonia
solution in the 0.05 liter buffer mix, the remaining volume must then be the ammonium chloride solution. So let (0.05 - x) represent the volume of the
NH4Cl solution:
$$[NH_4Cl]=\frac{(0.05~\mathrm{L}-x)(1.0~\mathrm{M})}{0.05~\mathrm{L}}$$
And now that we know the buffer's pH, volume and concentration of NH3 and NH4Cl, the only other thing we need is the
pKa of the acid. Once we look up that the pKa of NH4Cl is 9.25, we can then use the Henderson–Hasselbalch equation
to find the unknown volumes. Simply plug in all of the values and solve for x:
$$pH=p{K_{a}}+\log\left(\frac{[A^-]}{[HA]}\right)$$
$$pH=p{K_{a}}+\log\left(\frac{[NH_3]}{[NH_4Cl]}\right)$$
$$9.35=9.25+\log\left(\frac{\frac{x(0.70)}{0.05}}{\frac{(0.05-x)(1.0)}{0.05}}\right)$$
$$9.35=9.25+\log\left(\frac{x(0.70)}{(0.05-x)(1.0)}\right)$$
$$0.10=\log\left(\frac{0.70x}{0.05-x}\right)$$
$$10^{0.10}=\frac{0.70x}{0.05-x}$$
$$1.26=\frac{0.70x}{0.05-x}$$
$$0.063-1.26x=0.70x$$
$$0.063=1.96x$$
$$0.0321=x$$
Therefore, the volumes of 0.70 M NH3 and 1.0 M NH4Cl needed to prepare 50 mL of buffer with a pH of 9.35 are:
$$\mathrm{volume~of~NH_3}=x=0.0321~\mathrm{L}=32.1~\mathrm{mL}$$
$$\mathrm{volume~of~NH_4Cl}=0.05-x=0.05~\mathrm{L}-0.0321~\mathrm{L}=0.0179~\mathrm{L}=17.9~\mathrm{mL}$$
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