The MW of Metol is listed as 344.39, which includes 1 MW of H2SO4, 98.08. I used 1/8 MW of Metol, so 0.125 MW H2SO4 went in and were neutralized by 0.25 MW of KOH leaving 0.125 MW of K2SO4 and some water. I calculate now 22.01 g K2SO4. I must have slipped a cog sometime about 2 AM. I'm close to you, but not for the same reasons. There are TWO molecules of the Metol base for each molecule of H2SO4, so each MW of Metol requires 4 MW of KOH to completely free two MW of the base, which is (344.39 - 98.08)/2 or 123.16. Since I'm using 1/8 MW of Metol, I need 28 g KOH.