[clarkstill]

K2CO3  +  H2O  →  KOH + KHCO3
KHCO3  → KOH + CO2
Assuming that starting from anhydrous 283mg/138 = 2.05 mmol K2CO3, a 65% hydrolysis leads to:
0.65x2.05x56 = 74.6 mg KOH
0.65x2.05x100 = 133.3 mg KHCO3
(1-0.65)x2.05x138 = 99.0 mg K2CO3
Or 306.9 mg solid, which after thermal degradation of 1.1 mmoles of KHCO3 (83% conversion) leads to:
2.43x56 = 136.2 mg KOH
0.23x100 = 23.0 mg KHCO3
0.71x138 = 99.0 mg K2CO3
Or, 258.2 mg of solid containing about 53% KOH, per mass.
Note that the above example is indicative because other combinations of KOH, KHCO3 and K2CO3 can also give this mass value (258 mg).

PS: I hope not being wrong in calculations that I have never done since the high school.

What about pKa values? The first reaction you've drawn is an acid-base reaction in reverse: the pKa of NaHCO3 is about 10 (according to the Evans table), while the pKb of KOH is 14. So the reaction as you've drawn it will lie far to the left, with an equilibrium constant of 10^-4.

Can you cite any examples of the reactions you propose?

[rolnor]

My experiment shows that this "hydrolysis" clearly do not take place under these conditions.

[pgk]

1). I agree.
“rolnor” is right and I was wrong. Indeed KOH is soluble in warm ethanol. In this case, let’s assume that 258 mg of solid correspond to a mixture of KHCO3 and unhydrolyzed KHCO3. Thus:
100x + 138y = 258
x + y = 2.05
The above equations system gives that x = 0.66 mmoles KHCO3 and 1.4 mmoles K2CO3. Or, 0.66x56 = 37 mg KOH have been formed and removed by solvent decantation, which corresponds to a hydrolysis yield = 32%.
2). If well understood, anhydrous ethanol plus the stoichiometric amount of water were used. Thus:
2x2.05x18 = 74 mg H2O, which may help but it is unlike that can completely dissolve 283-258 = 25 mg K2CO3 in ethanol.
3). “clarkstill” is right too. This is an equilibrium reaction with an acid-base equilibrium constant = 4.3x10^-7 at room temperature that corresponds to a hydrolysis constant 10^-14/4.3x3x10^-7 = 2.3x10^-8, which means that the equilibrium is 4.3x10^-7/2.3x10^-8 = 5.3 folds favored to the left and which corresponds to a hydrolysis ratio (x + y = 1 and x = 5.3y) = 0.158 or a hydrolysis yield = 15.8% that can easily be doubled by heating, due to the temperature influence in the equilibrium constant.
4). But please, calm down. Everybody (and not chemists only) knows that NaOH is not massively produced during cleaning with aqueous soda ash (Na2CO3) or during cooking with baking soda (NaHCO3) and thus, all the above calculations (if not being incorrect) might be simple coincidences.
5). Regardless if the above calculations are wrong or not, the fact is that the hydrolysis is an equilibrium reaction (as “clarkstill” denotes) and therefore, the hydrolysis ratio of K2CO3 is completely different in presence of a benzyl chloride or an ester that reacts with KOH and continuously removes the equilibrium to the right.
6). In other words, a chemical reaction with a low yield (like this one), may not have a synthetic/preparative interest but as a side reaction, it can be more than enough to significantly decrease the reaction yield or even, it can completely minimize it.

[pgk]

CORRECTION
6). In other words, a chemical reaction with low yield (like this one), may not have a synthetic/preparative interest but as a side reaction, it can be more than enough to significantly decrease the yield of the ‘main’ reaction or even, it can completely minimize it.