# A 35.0mL sample of an unknown HClO_4 solution requires 50.3mL of 0.109M NaOH for complete neutralization. What was the concentration of the unknown HClO_4 solution? The neutralization reaction is: HClO_(4(aq))+NaOH_((aq))→H_2O_((l))+NaClO_(4(aq))

May 20, 2015

The molarity of the perchloric acid solution is equal to 0.166 M.

$H C l {O}_{4 \left(a q\right)} + N a O {H}_{\left(a q\right)} \to {H}_{2} {O}_{\left(l\right)} + N a C l {O}_{4 \left(a q\right)}$

Notice the $1 : 1$ mole ratio thatexists between the rectants. This tells you that a complete neutralization implies equal number of moles of perchloric acid and of sodium hydroxide.

Use the sodium hydroxide solution's molarity and volume to determine how many moles were needed to completely neutralize the acid

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

${n}_{N a O H} = \text{0.109 M" * 50.3 * 10^(-3)"L" = "0.00548 moles}$ $N a O H$

Automatically, this is equal to the number of moles of perchloric acid present in solution. This means that the molarity of the initial perchloric acid solution was

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

${C}_{H C l {O}_{4}} = \text{0.00548 moles"/(35.0 * 10^(-3)"L") = "0.1656 M}$

Rounded to three sig figs, the answer will be

${C}_{H C l {O}_{4}} = \textcolor{g r e e n}{\text{0.166 M}}$