Evaldas, if you look in this equation i suppose you must get the answer.

(0.01mol 'H2SO4'- 0.00193mol 'BaOH'- 1/2 mol KOH - 1/2 mol NaCl) + mol HCl = 0.00226 mol 'NaOH'

N.B. about H2SO4 / HCl it happens for sure but at the end all species are ionized " rule: strong acid can displace the weaker one from its salt" so it doesn't matter as 2H 'H2SO4' = 2H 'HCl' and the most important is that H+ is equivalent and present.

now back to the equation simply 1/2 mol NaCl = mol HCl because any NaCl converted to HCl will consume 1/2 mol of H2SO4.

N.B. in the above equation 1/2 in KOH, NaCl correspond to the consumed H2SO4 in the equations you posted by yourself.

so finally you left with

0.01mol 'H2SO4'- 0.00193mol 'BaOH'- 1/2 mol KOH = 0.00226 mol 'NaOH'

and from it " all mole are known except that of KOH" calculate moles of consumed H2SO4 for neutaralizing KOH and by multiplying it '2' you get the actual moles of KOH.