what volume (mL) of 0.0962M NaOH is required to neutralize 10.0mL of 0.128M H2SO4?

I got 26.6mL

I did, too.

I got 13.2mL

To find the volume of NaOH required to neutralize the H2SO4, we can use the concept of stoichiometry and the balanced equation for the neutralization reaction between NaOH and H2SO4.

The balanced equation for the reaction is:

2NaOH + H2SO4 -> Na2SO4 + 2H2O

From the balanced equation, we can see that 2 moles of NaOH react with 1 mole of H2SO4.

First, we calculate the moles of H2SO4 using the given concentration and volume:

Number of moles of H2SO4 = concentration x volume
= 0.128 M x 10.0 mL
= 0.00128 moles

Since the ratio of H2SO4 to NaOH is 1:2, we need twice the number of moles of NaOH to neutralize H2SO4.

Number of moles of NaOH = 2 x moles of H2SO4
= 2 x 0.00128 moles
= 0.00256 moles

Now, we can use the concentration of NaOH to find the volume needed to reach 0.00256 moles:

Volume of NaOH = Number of moles / Concentration
= 0.00256 moles / 0.0962 M
≈ 26.6 mL

Therefore, approximately 26.6 mL of 0.0962 M NaOH is required to neutralize 10.0 mL of 0.128 M H2SO4.