Determine how many milliliters of 0.102M NaOH solution is needed to neutralize 35.0mL of 0.125M H2 SO4 solution.

85.8ml

2NaOH + H2SO4 ==> Na2SO4 + 2H2O

moles H2SO4 = M x L = 0.125 x 0.035L = ??.
moles NaOH (look at the coefficients in the balanced equation) = 2 x moles H2SO4.
Finally, M = moles/L. Solve for L and convert to mL.

To determine the number of milliliters of the 0.102M NaOH solution needed to neutralize 35.0mL of the 0.125M H2SO4 solution, we can use the concept of stoichiometry.

First, let's write the balanced equation for the neutralization reaction between NaOH and H2SO4:

2 NaOH + H2SO4 -> Na2SO4 + 2 H2O

From the equation, we can see that it takes 2 moles of NaOH to neutralize 1 mole of H2SO4.

Now, let's calculate the number of moles of H2SO4 in the given 35.0mL of 0.125M H2SO4 solution:

Moles of H2SO4 = concentration (M) x volume (L)
= 0.125 mol/L x (35.0 mL / 1000 mL/ L)
= 0.125 mol/L x 0.0350 L
= 0.004375 mol

Since the stoichiometric ratio between H2SO4 and NaOH is 1:2, we need twice as many moles of NaOH to neutralize the H2SO4.

Moles of NaOH needed = 2 x Moles of H2SO4
= 2 x 0.004375 mol
= 0.00875 mol

Finally, let's calculate the volume of the 0.102M NaOH solution needed to contain 0.00875 moles:

Volume (L) = moles / concentration (M)
= 0.00875 mol / 0.102 mol/L
= 0.0858 L

To convert this volume to milliliters, we multiply by 1000:

Volume (mL) = 0.0858 L x 1000
= 85.8 mL

Therefore, you would need approximately 85.8 milliliters of the 0.102M NaOH solution to neutralize 35.0 milliliters of the 0.125M H2SO4 solution.

To determine how many milliliters of 0.102M NaOH solution is needed to neutralize 35.0mL of 0.125M H2SO4 solution, we can use the concept of stoichiometry.

First, we need to write a balanced chemical equation for the reaction between NaOH and H2SO4:

H2SO4 + 2NaOH -> Na2SO4 + 2H2O

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

Now, we can calculate the number of moles of H2SO4 in the 35.0mL of 0.125M H2SO4 solution:

moles of H2SO4 = volume (L) x concentration (M)
= 0.035 L x 0.125 M
= 0.004375 moles

Since the stoichiometry of the reaction is 1:2 (H2SO4:NaOH), we can determine that we need twice the number of moles of NaOH to react with the H2SO4:

moles of NaOH = 2 x moles of H2SO4
= 2 x 0.004375 moles
= 0.00875 moles

Now we can calculate the volume of 0.102M NaOH solution needed to provide 0.00875 moles of NaOH:

volume (L) = moles / concentration (M)
= 0.00875 moles / 0.102 M
≈ 0.0858 L

To convert this volume to milliliters, we can multiply by 1000:

volume (mL) = 0.0858 L x 1000
≈ 85.8 mL

Therefore, approximately 85.8 milliliters of 0.102M NaOH solution is needed to neutralize 35.0mL of 0.125M H2SO4 solution.