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NaOH & KOH titration procedure interchangeable?

November 5, 2012

Q. I have a simple colorimetric titration that we perform on our NaOH tank for determining concentration in oz/gal.

Could I use the same titration for a KOH tank that we may be installing?

Marc Green
anodizer - Boise, Idaho

November 6, 2012

A. The procedure will be the same. The fudge factor will be slightly different due to the wt of the Na ion vs the K ion.

James Watts
- Navarre, Florida

November 7, 2012

It will not be the same. Even though the molar ratios might be the same, the g/mol of KOH is higher than the g/mol of NaOH by a factor of 1.4.

For instance if you had 10 mL of a 1N (40 g/L) solution of NaOH and you titrated 1N (36.5 g/L) HCl against it, it would take 10.96 mL of HCl to neutralize all of the Hydroxide
Now, if you took 10 mL of a 1N (56.1 g/L) solution of KOH and you titrated 1N (36.5 g/L) HCl against it, it would take 15.37 mL of HCl to neutralize all of the Hydroxide.
10.96 x 1.4=15.344, which is pretty darn close to 15.37 (rescinded)

Scott Merritt
- Eastman, Georgia, USA

November 7, 2012

A. Good day Mark.
Yes, you could use the same titration, providing your factor for KOH is 1.415 x greater than for your NaOH factor. Hope this helps,
Regards,

Eric Bogner
- Toronto, Ontario, Canada

November 8, 2012

A. Hello,

If there is ONLY potassium and sodium hydroxide in each tank, the technique is the same but you must change the factor you use to get the concentration.

The new factor should be 1.403 times the old one (the one used for sodium hydroxide).

If there are other components aside the hydroxides, you should look for the right indicator to titrate only the potassium hydroxide in the new mix.

Hope this clarifies something! Best regards,

Daniel Montañés
- Cañuelas, Buenos Aires, Argentina

A. The titration procedure would remain the same since both KOH and NaOH are such strong bases. However, the calculation would change. V1 x N1 = V2 x N2, where V! is volume of titrant, N1 is normality of titrant, V2 is volume of sample, and N2 is normality of sample.

James Totter, CEF
- Tallahassee, Florida
November 8, 2012

November 10, 2012

A. Scott's answer is wrong. ml x N = ml x N. If the N is the same, then the ml has to be the same. Now the weight of the hydroxide to get a 1N solution will be different by the 1.4xx factor mentioned.
For a tank analysis, you do not know the N of the OH, so you multiply the ml of acid by an appropriate fudge factor to get the amount of whichever OH you are using.
His first sentence is correct.

James Watts
- Navarre, Florida

December 6, 2012

My apologies, its been a while since my college chem class and obviously I should be more careful.

Scott Merritt
- Fort Walton, Florida, USA

Q. Gentlemen, thank you very much. I consider myself very good at what I do, however chemistry is definitely a weak point for me (one of those things you think "yeah, I'll never use this crap in real world" while sitting in class). So I'd like to make sure I understand what you guys are telling me. My current titration for the NaOH is as follows:

2 ml of NaOH solution diluted to 5 ml with DI water
Couple drops of phenolphthalein [on eBay & Amazon]
Titrate to phenolphthalein end point with 0.1N HCl
ml of .1N HCl x 3 = concentration of NaOH in oz/gal

So, if I'm understanding correctly, instead of multiplying times 3, I should multiply times 4.2 (3 x 1.4) in order to get my concentration?

Marc Green
anodizer - Boise, Idaho
November 13, 2012

November 14, 2012

A. Good day Mark.
I have not seen/am familiar with your titration, but as long as you are confident with it (I always make known standards and analyze to verify a different/new analysis), continue.
I use a 10 ml sample with barium chloride, phenol, 0.1N HCl, filter, methyl orange [on eBay or Amazon] , 1.0N HCl.
This procedure gives me both OH & CO3 in one analysis for Na & K.
The factor for NaOH is 0.053 and the factor for KOH is 0.075.
Yes, the KOH factor is 1.415 times the factor for NaOH.
Oh, and yeah, I think I'm pretty good at what I do, yeah, I started out as a plater, "helped" out in waste treatment, and "helped" out in the lab, and now years later the lab is where I "help" out the platers!
Hope this helps.
Regards,

Eric Bogner
- Toronto, Ontario, Canada

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