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.

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.

See also Essential Principles of Organic Chemistry (1956)

It follows that a quaternary ammonium hydroxide is completely dissociated and should be a strong base, as is sodium hydroxide. The experimental facts are entirely consistent with this view. Although trimethylamine in water solution gives only a small concentration of hydroxide ion, tetramethylammonium hydroxide gives a full molar equivalent of hydroxide ion in solution.

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

Practically the preparation of tetramethylammonium hydroxide fiom 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 N, 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 TAMH 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.

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.