# Question e3465

Jun 14, 2015

The sulfuric acid solution had a concentration of 0.304 M.

#### Explanation:

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

Notice that you have a $1 : \textcolor{red}{2}$ mole ratio between sulfuric acid and potassium hydroxide. This means that, in order to have a complete neutralization, you're going to need twice as many moles of potassium hydroxide than of sulfuric acid.

Since molarity is defined as moles of solute per liters of solution, you can use the molarity and volume of the potassium hydroxide solution to determine how many moles of the base were needed to completely react with the acid.

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

${n}_{K O H} = 0.45 \text{mol"/cancel("L") * 50.0 * 10^(-3)cancel("L") = "0.0225 moles}$ $K O H$

This means that the sulfuric acid solution contained

0.0225cancel("moles"KOH) * ("1 mole "H_2SO_4)/(color(red)(2)cancel("moles"KOH)) = "0.01125 moles" ${H}_{2} S {O}_{4}$

Since you know the volume of the sulfuric acid and how many moles it contained, you can determine its molarity by

C = n/V = "0.01125 moles"/(37.0 * 10^(-3)"L") = color(green)("0.304 M")#