What volume of an aqueous solution with #"pH"# #9# can you use to neutralize an aqueous solution with #"pH"# #2#?

It really depends on the base you use. Without context, it's hard to tell.

If you have a pH of #2# for pure water spiked with strong monoprotic acid, then:

#-log["H"^(+)] = 2#

#["H"^(+)] = color(highlight)(10^(-2) "M")#

Let's say this acid turned out to be #"HCl"#.

If you have a pH of #9# for pure water spiked with strong monoprotic base, then:

#-log["H"^(+)] = 9#

#["H"^(+)] = 10^(-9) "M" => ["OH"^(-)] = color(green)(10^(-5) "M")#

If you want to use #"NaOH"# as your base, you need an equimolar amount, so:

#5 cancel("mL HCl") xx color(highlight)((10^(-2) "mol HCl")/(1000 cancel"mL HCl"))#

#= 5xx10^(-5) "mol HCl"#

#= color(blue)(5xx10^(-5) "mol H"^(+))#

The question is then, if you start with a solution of #"NaOH"# that gives you a pH of #9# by itself (that is, with #10^(-5) "M NaOH"#), then what volume of it do you need to equivocate the #"mols NaOH"# with the #"mols HCl"#?

To neutralize #5xx10^(-5) "mol H"^(+)#, you need #5xx10^(-5) "mol"# strong monoprotic base to give you #5xx10^(-5) "mol OH"^(-)#, in this case.

#[color(blue)(5xx10^(-5) "mol OH"^(-))]/("? mL NaOH") = color(green)((10^(-5) "mol OH"^(-))/("1000 mL NaOH") = 10^(-5) "M")#

You can plainly see then that if you scale the right side by #5#, you need #"5000 mL"# of #10^(-5) "M NaOH"# to neutralize #"5 mL"# of #10^(-2) "M HCl"#.

In this case, you need 1000x more base than you have acid, volume-wise.