A chemist uses a solution of 0.226 M (KOH) to titrate 89.07 mL (H2SO4), she finds that it requires 34.00 mL of the base to reach the end-point of the titration. What is the molarity of the acid solution ? must be rounded to the correct number of sig figs

The question basically tells us that #"34.00 mL"# of a #"0.226 M"# solution of #KOH# neutralised #"89.07 mL"# of an unknown concentration of #H_2SO_4#, assuming that end point equals neutralisation point.

Here's the balanced equation of the neutralisation reaction between #KOH# and #H_2SO_4#:

#2KOH + H_2SO_4 -> 2H_2O + K_2SO_4#

From this, we can tell that, for every #2# moles of #KOH# that reacts, #1# mole of #H_2SO_4# will react.
In other words, the number of moles of #H_2SO_4# will be half the number of moles of #KOH#.

Then, we need to find the number of #KOH# that reacted using this equation:

#"number of moles" = "concentration (M)" xx "volume (L)"#
#"number of moles" = "0.226 M" xx "34.00 mL"#
#"number of moles" = "0.226 M" xx "0.03400 L" = "0.00768 moles"#

Since the number of moles of #H_2SO_4# will be half the number of moles of #KOH#, it will be:

#("0.00768 moles " KOH)/2 = "0.00384 moles " H_2SO_4#

So, #"0.00384 moles"# will correspond to #"89.07 mL"# (which was given in the question), we can now find the concentration of #H_2SO_4#:

#"concentration (M)" = "number of moles"/"volume (L)"#

#"concentration (M)" = "0.00384 moles"/"89.07 mL"#

#"concentration (M)" = "0.00384 moles"/"0.08907 L" = "0.0431 M"#