If 60 ml of naoh solution neutralizes 40 ml of 0.50 m h2so4, the concentration of the naoh solution is most nearly:

The balanced chemical equation for this reaction is:

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

According to question, 60 ml of NaOH solution was used for neutralizing 40 ml of 0.50M H2SO4.

The no. of moles of H2SO4 is calculated using the equation:
mol solute = (molarity) (L soln)

mol H2SO4 = 0.50 M x 0.040 L = 0.02 moles of H2SO4

As per the equation, the number of moles of NaOH used is:

0.02 moles of H2SO4 (2 mol NaOH) (1 mol H2SO4) = 0.04 moles of NaOH

Therefore, using the given volume of NaOH, the concentration or molarity of NaOH can be calculated using the formula :

Molarity = mol solute/L soln = 0.04 mol NaOH/0.06 L = 0.67 M

Therefore, the concentration of NaOH is 0.67 M.

Because it was a dramatically big guy

Ithink 2 is the correct