# A basic question about tetramethylammonium hydroxide --strong or weak?

This is a follow-up to the question here. There is an answer claiming that tetramethylammonium hydroxide (TMAH) is a suitable strong base that can be used instead of the usual alkalies to reach pH 14. However, Wikipedia says that TMAH is only a weak base.

But, when I state this I get comments that other sources document TMAH as a strong base. The other sources, however, just mention it in passing and do not offer direct proof. Can anyone provide a reference that actually does settle this question? Thanks!

Note: At the time the question was posed Wikipedia did make the weak base claim. It has since been edited, and it now appears that the edit is correct. See the answers.

• Remark: Wikipedia backs its claim with this German thesis "Changes in high-molecular weight compounds during beech litter decomposition" as a reference. On printed page 12, the $\mathrm{p}K_\mathrm{b}$ is given to be 4.2. This citation was added to that WP page in Dec 2014 in this revision by user "Czeror". Might help in organizing clues to resolve the dilemna. – Gaurang Tandon Mar 17 '18 at 12:07
• Wikipedia clearly says it's a very strong base, I have no idea why you'd have doubts about that. Also title needs to be rewritten. – Mithoron Mar 17 '18 at 19:53
• @Mithoron the title is a great pun, I love it. Several journal articles also say 4.2, maybe the oldest being sciencedirect.com/science/article/pii/S0165237010000811 (2010) and then newer articles citing to that. Overall, I think it is a good question. – DavePhD Mar 17 '18 at 20:11
• @DavePhD Well, I feel title should be informative and clear in the first place, and then it may be also entertaining. – Mithoron Mar 17 '18 at 20:19

TETRAMETHYLAMMONIUM HYDROXIDE Journal of the Chemical Society 87 (1905): 955-961:

Practically the preparation of tetramethylammonium hydroxide from its salts resolves itself into a question of solubility as follows. In general, the equation

$$\ce{NMe4X + MOH = NMe4OH + MX}$$

will represent a real action proceeding nearly to completion if M, X, and the solvent are so chosen that all the substances except MX shall be soluble, or at least that MX shall be much less soluble than either of the original reacting substances. This principle was applied by Hofmann in his preparations with water as solvent. For M and X he chose either the pair Ag,I or Ba,SO4, in both of which cases MX is practically insoluble in water.

It is clear that if the general application of the above principle is justifiable, tetramethylammonium hydroxide may be prepared from a tetramethylammonium salt by means of potassium hydroxide if we so choose X and the solvent that of the substances represented in the equation

$$\ce{NMe4X + KOH = NMe4OH + KX}$$

all shall be soluble except KX.

...

An estimation of the strength of the base by means of the velocity of saponification of methyl acetate in N/80 solution showed that it was somewhat weaker than sodium hydroxide. The velocity constants obtained at 25 [degrees] were 0.0106 and 0.0115 respectively, so that if the strength of sodium hydroxide is represented as 100, that of tetramethylammonium hydroxide, under the above conditions, will be represented by 92.

So by two independent methods it is shown that tetramethylammonium hydroxide is much stronger than pKb=4.2. Firstly, the halide cannot not be converted to the hydroxide by adding metal hydroxide, unless an insoluble halide is produced to force the equilibrium forward. Seconding, it is shown by velocity of saponification to be only slightly weaker then NaOH.

DavePhD is right, and I am accepting and up-voting his answer. But first ... the real key to the conundrum is a little outside the box. It is trimethylamine that has a $\mathrm{p}K_\mathrm{b}$ of 4.2. Apparently some chemists mistook the basicity of TMAH for that of trimethylamine.