We're asked to find the volume (in #"mL"#) of a #0.45M# #"HCl soln"# that will neutralize #25.0# #"mL"# of a #1.00M# #"KOH soln"#.
To do this, let's first write the chemical equation for the reaction:
#"HCl"(aq) + "KOH"(aq) rarr "KCl"(aq) + "H"_2"O"(l)#
What we can do is use the molarity equation to find the number of moles of #"KOH"# that are present:
#"molarity" = "mol solute"/"L soln"#
#"mol solute" = ("molarity")("L soln")#
#"mol KOH" = (1.00"mol"/(cancel("L")))overbrace((0.0250cancel("L")))^"converted to liters" = color(red)(0.0250# #color(red)("mol KOH"#
Now, we can use the coefficients of the chemical equation to find the relative number of moles of #"HCl"#:
#color(red)(0.0250)cancel(color(red)("mol KOH"))((1color(white)(l)"mol HCl")/(1cancel("mol KOH"))) = color(red)(0.0250# #color(red)("mol HCl"#
Lastly, we can use the molarity equation again to find the volume of #"HCl"# solution:
#"molarity" = "mol solute"/"L soln"#
#"L soln" = "mol solute"/"molarity"#
#"L HCl soln" = (color(red)(0.0250)cancel(color(red)("mol HCl")))/(0.45(cancel("mol"))/"L") = color(blue)(0.056# #color(blue)("L"# #= color(blue)(ul(56color(white)(l)"mL"#
