# If K_a=2.0xx10^-11 for "hypoiodous acid", HOI(aq) what is [H_3O^+] for [HOI(aq)], whose concentration is 0.23*mol*L^-1?

Mar 4, 2017

$\left[{H}_{3} {O}^{+}\right]$ $=$ $2.1 \times {10}^{-} 6 \cdot m o l \cdot {L}^{-} 1$

#### Explanation:

We need (i) a stoichometrically balanced equation:

$H O I \left(a q\right) + {H}_{2} O \left(l\right) r i g h t \le f t h a r p \infty n s I {O}^{-} + {H}_{3} {O}^{+}$

And (ii), the equilibrium equation that represents the dissociation of the acid:

$\frac{\left[I {O}^{-}\right] \left[{H}_{3} {O}^{+}\right]}{\left[H O I\right]} = {K}_{a} = 2.0 \times {10}^{-} 11$

Now, using the typical approach we assume that the concentration of acid that dissociates is $x$, and thus we can rewrite the equilibrium equation,

$\frac{\left(x\right) \left(x\right)}{0.23 - x} = 2.0 \times {10}^{-} 11$

This is a quadratic in $x$, which is solvable exactly. Because I am a lazy soul, however, I make the approximation that $\left(0.23 - x\right) \cong 0.23$. Note that I must justify this approx. later.

So ${x}^{2} = 0.23 \times 2.0 \times {10}^{-} 11$,

${x}_{1} = \sqrt{0.23 \times 2.0 \times {10}^{-} 11} = 2.14 \times {10}^{-} 6$

This value is indeed small compared to $0.23$, but sometimes it is necessary to take a second approximation, or even a third approximation.

${x}_{2} = \sqrt{\left(0.23 - 2.14 \times {10}^{-} 6\right) \times 2.0 \times {10}^{-} 11} = 2.14 \times {10}^{-} 6$

Since ${x}_{1}$ and ${x}_{2}$ were identical, our approximation was correct, and our value is the same value that we would have obtained had we used the quadratic equation.

For some more examples of weak acid dissociation see here.