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 mole - ratio from balanced equation

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

Step3: Calculate molarity of NaOH

Use the formula $M=\frac{n}{V}$. We know $n_{NaOH} = 1\ mol$ and $V_{NaOH}=2.00\ L$. So $M_{NaOH}=\frac{1\ mol}{2.00\ L}=0.50\ M$.

Answer:

0.50 M