🤓 Based on our data, we think this question is relevant for Professor Goldsby's class at FSU.

We’re being asked **to calculate the [HClO _{2}]** at equilibrium when 1.0 L of a 0.35 M solution of HClO

Since HClO_{2 }is a weak acid it will not dissociate completely in the aqueous solution and NaClO.

Remember that * weak acids* partially dissociate in water and that

HClO_{2}(aq) ⇌ H^{+}(aq) + ClO_{2}^{–}(aq); **K _{a1} = 1.1 × 10^{–2}^{ }**(can be found in books and the internet)

ClO^{-} (aq) + H_{3}O+(aq) ⇌ HClO(aq) + H_{2}O (l); **Ka _{ }= 2.9 x 10^{-8} **(can be found in books and the internet)

The overall reaction is:

HClO_{2} (aq) + ClO- (aq) ⇌HClO (aq) + ClO^{2-} (aq)

From this, we can construct an ICE table. Remember that liquids are ignored in the ICE table.

The * K_{a} expression* for the reaction is:

$\overline{){\mathbf{K}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}{{\mathbf{K}}}_{{\mathbf{a}}}{{\mathbf{K}}}_{{\mathbf{b}}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}\frac{\mathbf{products}}{\mathbf{reactants}}{\mathbf{=}}{\mathbf{}}\frac{\mathbf{\left[}\mathbf{HClO}\mathbf{\right]}\mathbf{\left[}{\mathbf{NaClO}}_{\mathbf{2}}\mathbf{\right]}}{\mathbf{\left[}{\mathbf{HClO}}_{\mathbf{2}}\mathbf{\right]}\mathbf{\left[}\mathbf{NaClO}\mathbf{\right]}}}\u200a\phantom{\rule{0ex}{0ex}}$^{ }^{Liquids are ignored in the Ka expression.}

Note that each concentration is raised by the stoichiometric coefficient: [HClO_{2}], [NaClO2], [NaClO] and [HClO] are raised to 1.

To 1.0 L of a 0.35 M solution of HClO_{2} is added 0.18 mol of NaClO.

Calculate the [HClO_{2}] at equilibrium.