Assuming that pamoic acid is that soluble, here's how you would do the calculation.
Pamoic acid is a dibasic acid with the formula #"C"_23"H"_16"O"_6#.
We can write the formula as #"H"_2"C"_23"H"_14"O"_6#.
Then the equation for the neutralization is
#M_text(r):color(white)(m)338.375color(white)(mmm)39.997#
#color(white)(mm)"H"_2"C"_23"H"_14"O"_6 + "2NaOH" → "Na"_2"C"_23"H"_14"O"_6 + 2"H"_2"O"#
#color(white)(mmmml)"H"_2"A"color(white)(mll) + "2NaOH" → color(white)(mm)"Na"_2"A" color(white)(mll)+ 2"H"_2"O"#
Step 1. Calculate the moles of #"H"_2"A"#
#"Moles of H"_2"A" = 0.500 color(red)(cancel(color(black)("L H"_2"A"))) × ("1 mol H"_2"A")/(1 color(red)(cancel(color(black)("L H"_2"A")))) = "0.50 mol H"_2"A"#
Step 2. Calculate the moles of #"NaOH"#
#"Moles of NaOH" = 0.50 color(red)(cancel(color(black)("mol H"_2"A"))) × "2 mol NaOH"/(1 color(red)(cancel(color(black)("mol H"_2"A")))) = "1.0 mol NaOH"#
Step 3. Calculate the mass of #"NaOH"#
#"Mass of NaOH" = 1.0 color(red)(cancel(color(black)("mol NaOH"))) × "39.997 g NaOH"/(1 color(red)(cancel(color(black)("mol NaOH")))) = "40 g NaOH"#
