Respuesta :

To answer this question, we will need to use an ICE table.

So, we start with the concentration we want to calculate, [HClO].

The dissociation equilibrium is:

[tex]HClO\rightleftarrows H^++ClO^-[/tex]

At start, we only have HClO, so the first row of the table is:

HClO | H⁺ | ClO⁻

[HClO]i | 0 | 0

Now, suppose x dissociates at the equilibrium, so we would have:

HClO | H⁺ | ClO⁻

[HClO]i | 0 | 0

[HClO]i - x | x | x

Since the concentration of H⁺ will be x, we can calculate it by using the given pH:

[tex]\begin{gathered} pH=-\log \lbrack H^+\rbrack \\ x=\lbrack H^+\rbrack=10^{-pH}=10^{-3.2}\approx6.30957\times10^{-4}M \end{gathered}[/tex]

Now, we can figure the initial concentration of HClO by using the equilibrium equation:

[tex]Ka=\frac{\lbrack H^+\rbrack\lbrack ClO^-\rbrack}{\lbrack HClO\rbrack}[/tex]

From the ICE table, we have:

[tex]\begin{gathered} \lbrack H^+\rbrack=x\approx6.30957\times10^{-4}M \\ \lbrack ClO^-\rbrack=x\approx6.30957\times10^{-4}M \\ \lbrack HClO\rbrack=\lbrack HClO\rbrack_i-x=\lbrack HClO\rbrack_i-6.30957\times10^{-4}M \end{gathered}[/tex]

Thus:

[tex]3.0\times10^{-8}=\frac{6.30957\times10^{-4}M\cdot6.30957\times10^{-4}M}{\lbrack HClO\rbrack_i-6.30957\times10^{-4}M}[/tex]

Now, we can solve for [HClO]i:

[tex]\begin{gathered} 3.0\times10^{-8}=\frac{6.30957\times10^{-4}M\cdot6.30957\times10^{-4}M}{\lbrack HClO\rbrack_i-6.30957\times10^{-4}M} \\ 3.0\times10^{-8}(\lbrack HClO\rbrack_i-6.30957\times10^{-4}M)=3.98107\times10^{-7}M \\ 3.0\times10^{-8}\lbrack HClO\rbrack_i-1.89287\times10^{-11}=3.98107\times10^{-7}M \\ 3.0\times10^{-8}\lbrack HClO\rbrack_i=3.98107\times10^{-7}M+1.89287\times10^{-11}M \\ 3.0\times10^{-8}\lbrack HClO\rbrack_i=3.98126\times10^{-7}M \\ \lbrack HClO\rbrack_i=\frac{3.98126\times10^{-7}M}{3.0\times10^{-8}} \\ \lbrack HClO\rbrack_i\approx0.1327M \end{gathered}[/tex]

Thus, the initial concentration is approximately 0.1327 M.