You have the following species in solution.
From "H"_2"SO"_4:
0.0500 color(red)(cancel(color(black)("L H"_2"SO"_4))) × (0.100 color(red)(cancel(color(black)("mol H"_2"SO"_4))))/(1 color(red)(cancel(color(black)("L H"_2"SO"_4)))) × (2 "mol H"_3"O"^+)/(1 color(red)(cancel(color(black)("mol H"_2"SO"_4)))) = "0.0100 mol H"_3"O"^+
From "NaOH":
0.0250 color(red)(cancel(color(black)("L NaOH"))) × (0.200 color(red)(cancel(color(black)("mol NaOH"))))/(1 color(red)(cancel(color(black)("L NaOH")))) × (1 "mol OH"^"-")/(1 color(red)(cancel(color(black)("mol NaOH")))) = "0.005 00 mol OH"^"-"
From "Ba(OH)"_2:
0.0250 color(red)(cancel(color(black)("L Ba(OH)"_2))) × (0.100 color(red)(cancel(color(black)("mol NaOH"))))/(1 color(red)(cancel(color(black)("L NaOH")))) × ("2 mol OH"^"-")/(1 color(red)(cancel(color(black)("mol Ba(OH)"_2)))) = "0.005 00 mol OH"^"-"
From "KOH":
0.0100 color(red)(cancel(color(black)("L KOH"))) × (0.150 color(red)(cancel(color(black)("mol KOH"))))/(1 color(red)(cancel(color(black)("L KOH")))) × (1 "mol OH"^"-")/(1 color(red)(cancel(color(black)("mol KOH")))) = "0.001 50 mol OH"^"-"
From "HOCl":
0.0300 color(red)(cancel(color(black)("L HOCl"))) × (0.100 "mol HOCl")/(1 color(red)(cancel(color(black)("L HOCl")))) = "0.003 00 mol HOCl"
The 0.0100 mol of "H"_3"O"^+ from the "H"_2"SO"_4 will neutralize the total of 0.0100 mol of "OH"^"-" from the "NaOH" and "Ba(OH)"_2.
That leaves the reaction between "KOH" and "HOCl".
"HOCl" is a weak acid. It is partially neutralised by the "KOH".
We have generated a buffer.
color(white)(mmmmml)"HOCl" color(white)(ll)+ color(white)(m)"OH"^"-"color(white)(ll) → color(white)(m)"OCl"^"-" + "H"_2"O"
"I/mol:"color(white)(mll)"0.003 00"color(white)(ml)"0.001 50"color(white)(mmmm)0
"C/mol:"color(white)(m)"-0.001 50"color(white)(m)"-0.001 50"color(white)(m)"+0.001 50"
"E/mol:"color(white)(ml)"0.001 50"color(white)(mmml)0 color(white)(mmmll)"0.001 50"
For "HOCl", "p"K_"a" = 7.53.
According to the Henderson-Hasselbalch equation
"pH" = "p"K_"a" + log(("[OCl"^"-""]")/"[HOCl]") = 7.53 + log((stackrelcolor(blue)(1)(color(red)(cancel(color(black)("0.001 50"color(white)(l) "mol/L")))))/(color(red)(cancel(color(black)("0.001 50" color(white)(l)"mol/L"))))) = 7.53
