Question

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?

Answer 1

`1.64\ {mols}/{L}`, assuming that the molarity of `NaOH_{(aq)}` is `1.0\ {mols}/{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 `NaOH_{(aq)}=1.0\ mols`, 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.

`mols\ H_2SO_{4(aq)} * 2= mols\ NaOH_{(aq)}`

`10.0\ mL * {1\ L}/{1000\ mL} * {2\ mols\ NaOH_{(aq)}}/{1\ mol\ H_2SO_{4(aq)}} * x = 32.75\ mL * {1\ L}/{1000\ mL}*1.0\ {mols}/{L}`

`0.010\ L * {2\ mols\ NaOH_{(aq)}}/{1\ mol\ H_2SO_{4(aq)}} * x = 0.03275\ L * 1.0\ {mols}/{L}`

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

`x={0.03275\ L}/{0.010\ L} * {1\ mol\ H_2SO_{4(aq)}}/{2\ mols\ NaOH_{(aq)}} * 1.0\ {mols}/{L}`

After that, solve, and you have your answer of `1.6375\ {mols}/{L}`, or properly rounded, `1.64\ {mols}/{L}`.

Answer 2

The molarity of the `"H"_2"SO"_4` is 1.64 mol/L.

Explanation:

Step 1. Write the balanced chemical equation for the reaction

`"H"_2"SO"_4 + "2NaOH" → "Na"_2"SO"_4 + "2H"_2"O"`

Step 2. Calculate the moles of `"NaOH"`

I assume that the molarity of the `"NaOH"` is 1.00 mol/L.

`"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 `"H"_2"SO"_4`.

`"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 `"H"_2"SO"_4`

`"Molarity" = "moles"/"litres" = "0.016 375 mol"/"0.0100 L" = "1.64 mol/L"`