If the titration of a 30.0 mL sample of sulfuric acid requires 27.65 mL of 0.090 M sodium hydroxide, what is the molarity of the acid?

H2SO4(aq)+2NaOH(aq)→Na2SO4(aq)+2H2O(l)

mols NaOH = M x L = ?

Using the coefficients in the balanced equation convert mols NaOH to mols H2SO4.
Then M H2SO4 = mols H2SO4/L H2SO4 = ?

To find the molarity of the sulfuric acid, you can use the equation for the reaction between sulfuric acid and sodium hydroxide, and the volume and concentration of the sodium hydroxide solution used.

Given:
Volume of sulfuric acid = 30.0 mL
Volume of sodium hydroxide = 27.65 mL
Concentration of sodium hydroxide = 0.090 M

According to the balanced chemical equation:
H2SO4(aq) + 2NaOH(aq) → Na2SO4(aq) + 2H2O(l)

The ratio of sulfuric acid to sodium hydroxide is 1:2. This means that for every 1 mole of sulfuric acid, you need 2 moles of sodium hydroxide to completely react.

First, convert the volume of sodium hydroxide solution used to moles. To do this, you need to multiply the volume (in L) by the concentration (in mol/L):

Moles of sodium hydroxide = (27.65 mL) * (0.090 mol/L) / 1000 mL/L

Next, determine the moles of sulfuric acid from the moles of sodium hydroxide, using the mole ratio from the balanced chemical equation:

Moles of sulfuric acid = (moles of sodium hydroxide) / 2

Finally, calculate the molarity of the sulfuric acid:

Molarity of sulfuric acid = (moles of sulfuric acid) / (volume of sulfuric acid in L)

Substituting the values into the equation, we have:

Moles of sodium hydroxide = (27.65 mL) * (0.090 mol/L) / 1000 mL/L = 0.0024885 mol
Moles of sulfuric acid = 0.0024885 mol / 2 = 0.00124425 mol
Molarity of sulfuric acid = 0.00124425 mol / (30.0 mL / 1000 mL/L) = 0.0415 M

Therefore, the molarity of the sulfuric acid is 0.0415 M.