Sulfuric acid reacts with sodium hydroxide according to this equation:

H2S04 + 2 NaOH Na2(SO4) + 2 H2O

A 10.00 mL sample of the H2SO4 solution required 13.71 mL of 0.309 M NaOH for neutralization. Calculate the molarity of the acid.

0.01371L * 0.309 mol/L = 0.004236 mol NaOH

so, you need half that many moles of H2SO4, or 0.002118 mol H2SO4

0.002118 mol / 0.01L = 0.2118 M H2SO4

To calculate the molarity of sulfuric acid (H2SO4), we can use the balanced equation and the volume and concentration of sodium hydroxide (NaOH) used in the neutralization reaction.

From the balanced equation, we can see that the stoichiometric ratio between H2SO4 and NaOH is 1:2. This means that 1 mole of H2SO4 reacts with 2 moles of NaOH.

Given the volume of NaOH used (13.71 mL or 0.01371 L) and its concentration (0.309 M), we can calculate the number of moles of NaOH:

moles of NaOH = volume of NaOH (L) x concentration of NaOH (mol/L)
= 0.01371 L x 0.309 mol/L
= 0.00423139 mol

Using the stoichiometric ratio, we can determine the number of moles of H2SO4 that reacted:

moles of H2SO4 = (moles of NaOH) / 2
= 0.00423139 mol / 2
= 0.002115695 mol

Now, we can calculate the concentration (molarity) of the H2SO4 solution by dividing the number of moles of H2SO4 by the volume of the solution (10.00 mL or 0.01000 L):

molarity of H2SO4 = moles of H2SO4 / volume of H2SO4 (L)
= 0.002115695 mol / 0.01000 L
= 0.21157 M

Therefore, the molarity of the sulfuric acid solution is 0.21157 M.

To calculate the molarity of the sulfuric acid (H2SO4) solution, we need to use the stoichiometry of the reaction and the volume and concentration of the sodium hydroxide (NaOH) solution used for neutralization.

The balanced equation for the reaction between sulfuric acid and sodium hydroxide is:

H2SO4 + 2 NaOH -> Na2SO4 + 2 H2O

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

Given:

Volume of NaOH solution (V2) = 13.71 mL = 0.01371 L
Concentration of NaOH solution (C2) = 0.309 M

First, let's find the number of moles of NaOH used:

Moles of NaOH (n2) = Volume (V2) x Concentration (C2)
= 0.01371 L x 0.309 mol/L

Next, we can use the stoichiometry of the reaction to find the number of moles of H2SO4:

Moles of H2SO4 (n1) = Moles of NaOH (n2) / Stoichiometric ratio
= n2 / 2

Finally, to find the molarity of the H2SO4 solution:

Molarity of H2SO4 (C1) = Moles of H2SO4 (n1) / Volume of H2SO4 solution (V1)
= n1 / 0.010 L

Since the volume of the H2SO4 solution is not given, we'll assume it's 10.00 mL = 0.010 L.

Plugging in the values, we get:

Molarity of H2SO4 (C1) = (n2 / 2) / 0.010 L

Calculating all the values and substituting them into the equation, we can determine the molarity of the H2SO4 solution.