a chemist uses 0.25 l of 2.00 m h2so4 to completely neutralize a 2.00 l of solution of naoh. the balanced…

Explanation:

Step1: Calculate moles of H₂SO₄

Use the formula $n = M\times V$, where $n$ is moles, $M$ is molarity and $V$ is volume. For H₂SO₄, $M = 2.00\ M$ and $V=0.25\ L$. So $n_{H_2SO_4}=2.00\ mol/L\times0.25\ L = 0.5\ mol$.

Step2: Determine moles of NaOH from mole - ratio

From the balanced equation $2NaOH + H_2SO_4\rightarrow Na_2SO_4 + 2H_2O$, the mole - ratio of $NaOH$ to $H_2SO_4$ is 2:1. So $n_{NaOH}=2\times n_{H_2SO_4}$. Substituting $n_{H_2SO_4}=0.5\ mol$, we get $n_{NaOH}=2\times0.5\ mol = 1.0\ mol$.

Step3: Calculate molarity of NaOH

Use the formula $M=\frac{n}{V}$. We know $n_{NaOH} = 1.0\ mol$ and $V_{NaOH}=2.00\ L$. So $M_{NaOH}=\frac{1.0\ mol}{2.00\ L}=0.50\ M$.

Answer:

0.50 M