posted by Rebekah .
Determine the volume of 0.240 M KOH solution required to neutralize each of the following samples of sulfuric acid. The neutralization reaction is shown below.
H2SO4(aq) + 2 KOH(aq)---> K2SO4(aq) + 2 H2O(l)
a) 26 mL of 0.240 M H2SO4
b) 143 mL of 0.14 M H2SO4
for a I got .012 and for b I got .17 but they are both wrong:(
I don't know how you came up with those numbers.
2KOH + H2SO4 ==> K2SO4 + 2H2O
moles H2SO4 = M x L = ??
Use the coefficients to change moles H2SO4 to mole KOH. You can see that moles NaOH = 2x moles H2SO4
Then MNaOH = moles NaOH/L NaOH
Then convert L NaOH to mL.
You should obtain 52 mL.
Im confused, where did you get NaOH?