NaOH + HCl ---> NaCl + H2O
a --------------> a
KOH + HCl ----> KCl + H2O
b ---------------> b
Ta có hpt :
$\left \{ {{40a +56 b=6,08} \atop {58,5a + 74,5b=8,3}} \right.$
⇔$\left \{ {{a=0,04} \atop {b=0,08}} \right.$
m$m_{NaOH}$ = 0,04.40=1,6 (g)
$m_{KOH}$ = 0,08.56 =4,48 (g)