Helpppppppppppp lol

Question

A 50.00 mL sample of a NaOH solution is titrated with a 1.605 M solution of H2SO4. The titration requires 24.09 mL of the acid solution to reach the equivalence point. What is the molarity of the base solution?

Answer 1

Short answer: the NaOH is 1.547 M.

The whole problem turns on one thing: sulfuric acid is diprotic, so the mole ratio here is one acid to two base, not one to one.

H2SO4 + 2 NaOH -> Na2SO4 + 2 H2O

Step 1. Moles of acid delivered at the equivalence point.

mol H2SO4 = M x V
          = 1.605 mol/L x 0.02409 L
          = 0.0386645 mol

Step 2. Convert to moles of base using the 1 to 2 ratio.

mol NaOH = 2 x 0.0386645 mol
         = 0.0773289 mol

Step 3. Divide by the volume of the base sample.

M NaOH = 0.0773289 mol / 0.05000 L
       = 1.546578 mol/L
       = 1.547 M   (four significant figures)

Where this problem is usually lost: reaching straight for M1V1 = M2V2. That shortcut quietly assumes a one to one reaction, and here it hands you 0.7733 M, exactly half the right answer. If you want a one-liner that is actually safe for titrations, carry the proton count n:

n(acid) x M(acid) x V(acid) = n(base) x M(base) x V(base)
2 x 1.605 x 24.09 = 1 x M x 50.00
M = 77.3289 / 50.00 = 1.547 M

n is 2 for H2SO4 because it donates two protons, and 1 for NaOH because it accepts one. Write the balanced equation out before you touch the calculator and that factor of 2 never goes missing.