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?

2 Answers
Jun 15, 2016

Answer:

#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}#.

Aug 7, 2016

Answer:

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"#