When titrating 50.0 mL of 0.10 M H2SO4 with 0.10 M NaOH, how many mL of NaOH will you have added to reach the 1st equivalence point?

To determine the number of milliliters (mL) of NaOH needed to reach the first equivalence point when titrating 50.0 mL of 0.10 M H2SO4 with 0.10 M NaOH, we need to consider the stoichiometry of the reaction.

The balanced chemical equation for the reaction between H2SO4 and NaOH is:

H2SO4 + 2NaOH -> Na2SO4 + 2H2O

From the balanced equation, we can see that 1 mole of H2SO4 reacts with 2 moles of NaOH. This means that to completely react with 1 mole of H2SO4, we need 2 moles of NaOH.

First, let's calculate the number of moles of H2SO4 in 50.0 mL (0.0500 L):

moles of H2SO4 = volume (L) x concentration (M)
= 0.0500 L x 0.10 M
= 0.00500 moles

Since the ratio of H2SO4 to NaOH in the balanced equation is 1:2, we need twice the number of moles of NaOH to react with H2SO4.

moles of NaOH needed = 2 x moles of H2SO4
= 2 x 0.00500 moles
= 0.0100 moles

Now, let's calculate the volume of 0.10 M NaOH needed to obtain 0.0100 moles:

volume (L) = moles / concentration (M)
= 0.0100 moles / 0.10 M
= 0.100 L
= 100 mL

Therefore, to reach the first equivalence point, you would need to add 100 mL of 0.10 M NaOH to the 50.0 mL of 0.10 M H2SO4.

Refer to the H2A problem above BUT you only titrate the first H on H2SO4. Post your work if you get stuck.