# Homework Help: Raising the pH of a Solution

1. May 2, 2016

### Justin LaRose

1. The problem statement, all variables and given/known data

I work in a lab, normally we evaporate our acid and dilute with water, but I have 300 samples and this would destroy the fume hood so I need to raise the pH using ammonia.

known data:

Samples are 40ml of 9M HCl
I need to add Xml of XM ammonium hydroxide (aka aqueous ammonia)
The desired final pH needs to be between 0.8 and 2.

2. Relevant equations

First you need the reaction of HCl + NH3--> NH4+ + Cl-

we have n = (.04L)(9mol/L) = 0.36 mol HCl

The molarity of aqueous ammonia is about 14.5M but I think it cannot be exactly determined.

Once we go through this equation with the known moles of ammonia, we can subtract that amount from the reactants and add it to the products

THen we use the equation HCl + H20 --> H30+ + Cl-

using a Ka = Kw/Kb = the concentration of products over reactants we get

5.56x10^-10 = [H30+][Cl-]/[HCl]

where [H3O+] is x, we solve for x, then take the negative log to get the pH.

3. The attempt at a solution

I completed the problem based on the protocol written above and got 2 microliters of 0.001M NH3 to obtain a pH of ~1. However, when I tried to neutralize a test sample in the lab, it took about 21ml of concentrated ammonia to raise the pH of 40ml of HCl to about 0.4. I am not sure if I am doing something wrong with the calculation or with the practice. I did think that perhaps the immense amount of vapors produced from adding concentrated NH3 (aq) to 9M HCl is reducing the efficiency of the buffer. Perhaps if I try to do this with the diluted NH3 it will work. I will try this tomorrow but in the meantime, can anyone verify this calculation or steer me in the right direction? Thanks.

Last edited by a moderator: May 2, 2016
2. May 2, 2016

### Charles Link

I think this might be a simple acid -base problem. 40 ml of 9M HCl makes for .36 moles of H+. Assuming you have 14.5 M NH3, the number of moles of OH- is $x=14.5 V_x$. Solving for H+ concentration in the mixture $(.36-14.5V_x)/(.040+V_x)=.1$ if the final pH is 1.0. For pH=2, set to .01. It would appear the solution could go very quickly from acidic to basic. You might want to try partially neutralizing, and then use a diluted ammonia solution, and test with pH indicators, etc. It's been quite a number of years since I worked an acid-base problem, but I think I did it correctly. Once the numerator goes negative, you need to reverse it, and it represents the OH concentration. [H+][OH-]=10^-14 etc. I solved for $V_x$ above and got $V_x=25 ml$ (approximately). A word of caution: I would be very careful about mixing a strong acid and a strong base. I do think it could be rather explosive.

3. May 2, 2016

### Staff: Mentor

Thread closed temporarily for safety Moderation...

4. May 3, 2016

### Staff: Mentor

Are you sure you know what you are doing, and that adding ammonia won't destroy the samples? Ar you sure evaporation of the acid is a method of raising the pH? Getting rid of the acid is a standard lab procedure in analysis of ore/metals, but adding ammonia can make things messy. Not only you introduce another substance, you also create a salt that probably should not be there, and ammonia can react with things that are present in the samples.

And, as Charles Link wrote, be careful. These are concentrated solutions, so the reaction will be highly exothermic, samples can start to boil and splash if you will mix reagents too fast.

Last edited: May 3, 2016
5. May 3, 2016

### Justin LaRose

She does know what she is doing. This is my best friend, she's the smartest person at Wayne State. She wanted me to thank you for your advice, Charles, she said she was going to attempt this before.

Please stop telling me someone is going to be blowing something up. This is a professional scientist with several publications. She's using state of the art equipment. She didn't want to ask her doctoral advisor and was having trouble with this one problem, that's all. I forgot chemistry, I'm a physics major, I couldn't help her. This is just about the chemistry and figuring out why the theoretically derived numbers aren't working in practice.

6. May 3, 2016

### Charles Link

The calculation I did in post #2 is a simple one and it looks like it agrees with the V= 21 ml that you found experimentally. And I also like Borek's comments to be careful. Ammonia, is not something you want to have released all over the lab. (In a high school chemistry class, my lab partner handed a test tube of ammonia solution to me and said, "smell this, it makes good smelling salts", and I took a whiff before I realized what he said...it gave a good sting to the inside of my nose...)

7. May 3, 2016

### Staff: Mentor

Would make a perfect "Last words of...".

8. May 3, 2016

### Charles Link

Just to add to my comments about stinging the nose, both HCl and NH3 in these concentrations, if they start spewing out upon mixing, they can easily cause severe burns. In this case, much better to be extra careful...

9. May 3, 2016

### Justin LaRose

"Tell Charles:

Charles, I am encouraged by the fact that your calculation matches my experimental work. It has been so long since I've done freshman chemistry that I embarrassingly feel unfamiliar with your approach. I see how your calculation worked, but could you explain how you were able to set it up, based on the original reaction?

Also, to address the danger-level of the work, yes it does boil and fume and spit, but I add the ammonia dropwise to account for this and have the appropriate PPE on. Thank you for your concerns, safety first!!"

10. May 3, 2016

### Justin LaRose

"I studied the calculation for a minute or so longer and I understand how the equation is set up, so I no longer need an explanation from Charles unless he wants to explain it.

Thanks all (thank you Justin as well)!"

Thank you fellows very much for the help here. This friend of mine rarely asks for any type of help so it's much appreciated, she has solved her problem in the lab just now and there was no explosion :)

I hope you all have a good day or night wherever you are in space!

11. May 3, 2016

### Charles Link

Glad you got it figured out. For anyone else who is interested in how it works, the numerator is a one-for-one neutralization of the H+ by the OH- so that the number of moles of H+ remaining in the solution will be .36 minus the number of moles of OH- that are introduced. The denominator is the volume of the combined solutions in liters. The ratio is the concentration of H+ in moles per liter. Meanwhile pH=-log (H+ concentration). The equation can readily be solved for the necessary volume $V_x$ of NH3 solution needed (in liters ) to produce the selected remaining H+ concentration of the mixture.(Note the numerator of our formula only works if it gives H+>0. If it gives H+<0, the absolute value of the formula is then the OH- concentration in moles/liter, and the H+ concentration could then be found by [H+]=(+1.0 E-14)/[OH-]. And if the moles of H+ in the formula is very nearly zero (acid precisely neutralized), then [H+]=[OH-]=+1.0 E-7 moles/liter and pH=7, the same as for ordinary water. (The entire pH range from 4 to 10 is relatively neutral, and if H+ in the original formula is near zero, the pH is somewhere in that range.)) For pH=1, [H+]=.1 (moles/liter) . For pH=2, [H+]=.01 (moles/liter). And very glad to hear you took the necessary precautions, and successfully neutralized the acid.

Last edited: May 3, 2016
12. May 4, 2016

### Charles Link

Just as a follow-up to this one, when needing to neutralize a strong acid or a strong base, I would recommend first diluting the solutions by pouring the strong acid and/or strong base into water (carefully of course) and getting each down to a strength of 1M (1 molar) or thereabouts before mixing them together. At 1M, they aren't nearly as concentrated and there is a lot more water to absorb any heat that is given off upon mixing.

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