Sciencemadness Discussion Board - How to make a buffer of a specific pH?

How to make a buffer of a specific pH?

bolbol - 8-10-2016 at 11:22

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?

aga - 9-10-2016 at 11:18

pH means the negative log of the concentration of H3O+ in a solution.

-log(x) = 9.35, so 10-9.35 is the concentration of H3O+ 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-

= 4.65

so the conc of OH- = 10 -4.65

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 ....

Quote: Originally posted by bolbol

I did the math and I found the ratio of NH3/NH4+ required for that solution

NH₃ is NH₄+ in aqueous solution.

Oh dear, i'm lost now. Let the whipping begin.

[Edited on 9-10-2016 by aga]

DistractionGrating - 9-10-2016 at 13:55

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]

Darkstar - 10-10-2016 at 03:49

Quote: Originally posted by aga

Oh crap. Always read the question at least 5 times when drunk ....

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 pK_a of the acid. Once we look up that the pK_a 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:

$$\mathrm{volume~of~NH_3}=x=0.0321~\mathrm{L}=32.1~\mathrm{mL}$$
$$\mathrm{volume~of~NH_4Cl}=0.05-x=0.05~\mathrm{L}-0.0321~\mathrm{L}=0.0179~\mathrm{L}=17.9~\mathrm{mL}$$