# Consider the titration of 50 mL of 0.0200 M HCLO(aq) with 0.100 M NaOH (aq). What is the formula of the main species in the solution after the addition of 10.0 mL of base?

Aug 10, 2016

${\text{ClO}}^{-}$

#### Explanation:

The idea here is that you need to figure out if you're dealing with a complete neutralization, in which case you can say right from the start that the main chemical species present in solution after the reaction takes place will be the conjugate base of the weak acid.

You're mixing hypochlorous acid, $\text{HClO}$, a weak acid, with sodium hydroxide, $\text{NaOH}$, a strong base. As you know ,the strong base dissociates completely in aqueous solution to produce hydroxide anions, ${\text{OH}}^{-}$.

The balanced chemical equation that describes this reaction looks like this

${\text{HClO"_ ((aq)) + overbrace("OH"_ ((aq))^(-))^(color(blue)("coming from NaOH")) -> "ClO"_ ((aq))^(-) + "H"_ 2"O}}_{\left(l\right)}$

Notice that the two reactants are consumed in a $1 : 1$ mole ratio; moreover, the reaction produces $1$ mole of hypochlorite anions, ${\text{ClO}}^{-}$, for every $1$ mole of hypochlorous acid and $1$ moles of hydroxide anions consumed.

Use the molarities and volumes of the two solutions to find the number of moles of hypochlorous acid and sodium hydroxide, i.e. hydroxide anions -- do not forget that the volume must be expressed in liters!

50.0 * 10^(-3) color(red)(cancel(color(black)("L"))) * "0.0200 moles HClO"/(1color(red)(cancel(color(black)("L")))) = 10^(-3)"moles HClO"

10.0 * 10^(-3) color(red)(cancel(color(black)("L"))) * "0.100 mol OH"^(-)/(1color(red)(cancel(color(black)("L")))) = 10^(-3)"moles OH"^(-)

As you can see, you have equal numbers of moles of hypochlorous acid and of hydroxide anions. This means that both reactants will be completely consumed by the reaction.

As a result, ${10}^{- 3}$ moles of hypochlorite anions will be produced.

You can thus say that the main chemical species present in the resulting solution, if you exclude water, of course, will be the hypochlorite anion, ${\text{ClO}}^{-}$.

The hypochlorite anion will act as a base and accepts a proton from water to reform some of the hypochlorous acid and produce hydroxide anions, which is why the pH of the resulting solution will be higher than $7$, i.e. the solution will be basic.