The equation for the reaction is
#"H"_3"O"^"+" + "OH"^"-" → "2H"_2"O"#
1. Calculate the concentration of #"H"_3"O"^"+"#
#"pH = 1.097"#
#["H"_3"O"^"+"] = 10^"-pH"color(white)(l) "mol/L" = 10^"-1.097"color(white)(l)"mol/L" = "0.0800 mol/L" = "0.0800 mmol/mL"#
2. Calculate the moles of #"H"_3"O"^"+"#
#"Moles of H"_3"O"^"+" = 10.00 color(red)(cancel(color(black)("mL H"_3"O"^"+"))) ×( "0.0800 mmol H"_3"O"^"+")/(1 color(red)(cancel(color(black)("mL H"_3"O"^"+")))) = "0.800 mmol H"_3"O"^"+""#
3. Calculate the moles of #"OH"^"-"# needed
#"Moles of OH"^"-" = 0.800 color(red)(cancel(color(black)("mmol H"_3"O"^"+"))) × ("1 mmol OH"^"-")/(1 color(red)(cancel(color(black)("mmol H"_3"O"^"+")))) = "0.800 mmol OH"^"-"#
4. Calculate the concentration of #"OH"^"-"#
#"pH = 13.00"#
#"pOH = 14.0 - pH = 14.00 - 13 00 = 1.00"#
#["OH"^"-"] = 10^"-pOH"color(white)(l) "mol/L" = 10^"-1.00"color(white)(l)"mol/L" = "0.100 mol/L" = "0.100 mmol/mL"#
5. Calculate the volume of the #"OH"^"-"#
#"Volume of OH"^"-" = 0.800 color(red)(cancel(color(black)("mmol OH"^"-"))) × ("1 mL OH"^"-")/(0.100 color(red)(cancel(color(black)("mmol OH"^"-")))) = "8.00 mL OH"^"-"#
The titration will require 8.00 mL of #"NaOH"#.