# Question #f131c

Apr 22, 2015

The concentration of the hydronium ions will be 0.0199 M.

So, you're titrating perchloric acid, $H C l {O}_{4}$, which is a strong acid, with potassium hydroxide, $K O H$, a strong base.

The balanced chemical equation for the neutralization reaction that takes place is

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

Notice that you have a $1 : 1$ mole ratio between perchloric acid and potassium hydroxide, which means that a complete neutralization would require equal number of moles of each reactant.

If you have more moles of perchloric acid than you have of potassium hydroxide, the solution will still be acidic. If it's the other way around, the solution will be basic.

Use the volumes and molarities of the two solutions to determine how many moles of each you add together

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

${n}_{H C l {O}_{4}} = 25.0 \cdot {10}^{- 3} \text{L" * "0.723 M" = "0.018075 moles }$ $H C l {O}_{4}$

${n}_{K O H} = 60.0 \cdot {10}^{- 3} \text{L" * "0.273 M" = "0.01638 moles KOH}$

The substance with the fewer moles present in solution will be consumed. At the same time, the substance that remains will have fewer moles present in solution.

Since you have more acid than base present, you'll get

${n}_{H C l {O}_{4}} = 0.018075 - 0.01638 = \text{0.001695 moles}$

${n}_{K O H} = 0$

Since perchloric acid is a strong acid, it will dissociate completely in aqueous solution to give ${H}_{3} {O}^{+}$ and $C l {O}_{4}^{-}$ ions.

$H C l {O}_{4 \left(a q\right)} + {H}_{2} {O}_{\left(l\right)} \to {H}_{3} {O}_{\left(l\right)}^{+} + C l {O}_{4 \left(a q\right)}^{-}$

Since 1 mole of $H C l {O}_{4}$ produces 1 mole of hydronium ions, the number of moles of ${H}_{3} {O}^{+}$ will be

${n}_{{H}_{3} {O}^{+}} = {n}_{H C l {O}_{4}} = 0.001695$

The volume of the solution will now be

${V}_{\text{sol}} = {V}_{H C l {O}_{4}} + {V}_{K O H}$

${V}_{\text{sol" = 25.0 + 60.0 = "85.0 mL}}$

Therefore, the concentration of hydronium ions will be

$\left[{H}_{3} {O}^{+}\right] = \text{0.001695 moles"/(80.0 * 10^(-3)"L") = "0.01994 M}$

Rounded to three sig figs, the answer will be

$\left[{H}_{3} {O}^{+}\right] = \textcolor{g r e e n}{\text{0.0199 M}}$