# A 10.0 mL sample of H_2SO_4 solution from an automobile battery requires 32.75 mL of M NaOH for a complete reaction. What is the molarity of the H_2SO_4 solution?

Jun 15, 2016

$1.64 \setminus \frac{m o l s}{L}$, assuming that the molarity of $N a O {H}_{\left(a q\right)}$ is $1.0 \setminus \frac{m o l s}{L}$.

#### Explanation:

Since both solutions are a strong acid/base, no ${K}_{a} / {K}_{b}$ calculations are required. In a reaction between sulfuric acid and sodium hydroxide, two mols of base are required to fully neutralize one mol of acid.

Also, for the purposes of this question, I will assume that the molarity of $N a O {H}_{\left(a q\right)} = 1.0 \setminus m o l s$, since this information was accidentaly skipped in the question. (Or defaults to 1.0 mols/L anyways, as Michael suggested)

Step 1) Write the equation.

$m o l s \setminus {H}_{2} S {O}_{4 \left(a q\right)} \cdot 2 = m o l s \setminus N a O {H}_{\left(a q\right)}$

$10.0 \setminus m L \cdot \frac{1 \setminus L}{1000 \setminus m L} \cdot \frac{2 \setminus m o l s \setminus N a O {H}_{\left(a q\right)}}{1 \setminus m o l \setminus {H}_{2} S {O}_{4 \left(a q\right)}} \cdot x = 32.75 \setminus m L \cdot \frac{1 \setminus L}{1000 \setminus m L} \cdot 1.0 \setminus \frac{m o l s}{L}$

$0.010 \setminus L \cdot \frac{2 \setminus m o l s \setminus N a O {H}_{\left(a q\right)}}{1 \setminus m o l \setminus {H}_{2} S {O}_{4 \left(a q\right)}} \cdot x = 0.03275 \setminus L \cdot 1.0 \setminus \frac{m o l s}{L}$

Step 2) Isolate the variable being solved for. ($x$)

$x = \frac{0.03275 \setminus L}{0.010 \setminus L} \cdot \frac{1 \setminus m o l \setminus {H}_{2} S {O}_{4 \left(a q\right)}}{2 \setminus m o l s \setminus N a O {H}_{\left(a q\right)}} \cdot 1.0 \setminus \frac{m o l s}{L}$

After that, solve, and you have your answer of $1.6375 \setminus \frac{m o l s}{L}$, or properly rounded, $1.64 \setminus \frac{m o l s}{L}$.

Aug 7, 2016

The molarity of the ${\text{H"_2"SO}}_{4}$ is 1.64 mol/L.

#### Explanation:

Step 1. Write the balanced chemical equation for the reaction

$\text{H"_2"SO"_4 + "2NaOH" → "Na"_2"SO"_4 + "2H"_2"O}$

Step 2. Calculate the moles of $\text{NaOH}$

I assume that the molarity of the $\text{NaOH}$ is 1.00 mol/L.

$\text{Moles of NaOH" = "0.032 75" color(red)(cancel(color(black)("L NaOH"))) × "1.00 mol NaOH"/(1 color(red)(cancel(color(black)("L NaOH")))) = "0.032 75 mol NaOH}$

Step 3. Calculate the moles of ${\text{H"_2"SO}}_{4}$.

${\text{Moles of H"_2"SO"_4 = "0.032 75" color(red)(cancel(color(black)("mol NaOH"))) × ("1 mol H"_2"SO"_4)/(2 color(red)(cancel(color(black)("mol NaOH")))) = "0.016 375 mol H"_2"SO}}_{4}$

Step 4. Calculate the molarity of the ${\text{H"_2"SO}}_{4}$

$\text{Molarity" = "moles"/"litres" = "0.016 375 mol"/"0.0100 L" = "1.64 mol/L}$