# Question efddd

Apr 22, 2015

You need 21.8 mL of your sulfuric acid solution to neutralize that much sodium hydroxide.

${H}_{2} S {O}_{4 \left(a q\right)} + \textcolor{red}{2} N a O {H}_{\left(a q\right)} \to N {a}_{2} S {O}_{4 \left(A q\right)} + 2 {H}_{2} {O}_{\left(l\right)}$

Notice that you have a $1 : \textcolor{red}{2}$ mole ratio between sulfuric acid and sodium hydroxide. This means that you need 2 moles of the lattter for every 1 mole of the former in order for complete neutralization to take place.

If you have less than 1 mole of sulfuric acid for every 2 moles of sodium hydroxide, you won't get complete neutralization, i.e. you'll have base remaining in solution.

Use sodium hydroxide's molar mass to determine how many moles of $N a O H$ must be neutralized

0.220cancel("g") * "1 mole NaOH"/(39.997cancel("g")) = "0.00550 moles NaOH"

This means that you'd need

0.00550cancel("moles NaOH") * ("1 mole "H_2SO_4)/(color(red)(2)cancel("moles NaOH")) = "0.00275 moles"  ${H}_{2} S {O}_{4}$

Since you're dealing with a sulfuric acid solution, you can use its molarity to determine what volume would contain this many moles

$C = \frac{n}{V} \implies V = \frac{n}{C}$

V = (0.00275cancel("moles"))/(0.126cancel("moles")/"L") = "0.02183 L"#

Expressed in mL and rounded to three sig figs, the answer will be

$V = \textcolor{g r e e n}{\text{21.8 mL}}$