# The Ka of a monoprotic weak acid is 2.08 xx 10^-3. What is the percent ionization of a .194 M solution of this acid?

Apr 6, 2017

$\text{% ionization}$ $\cong$ 10%

#### Explanation:

We interrogate the equilibrium:

$H A + {H}_{2} O r i g h t \le f t h a r p \infty n s {H}_{3} {O}^{+} + {A}^{-}$

For which ${K}_{a} = 2.08 \times {10}^{-} 3$.

i.e. ${K}_{a} = \frac{\left[{H}_{3} {O}^{+}\right] \left[{A}^{-}\right]}{\left[H A\right]}$, and if $x$ moles of the initial concentration dissociates, then we have.............

$2.08 \times {10}^{-} 3 = \frac{{x}^{2}}{0.194 - x}$,

this is a quadratic in $x$, the which we could solve exactly. But because chemists are lazy folk, we make the approx. that $0.194 - x \cong 0.194$, we must justify this approx. later.

And so $2.08 \times {10}^{-} 3 = \frac{{x}^{2}}{0.194}$

i.e. ${x}_{1} = \sqrt{2.08 \times {10}^{-} 3 \times 0.194} = 0.020$, this value is indeed small compared to $0.194$, but now we have an approximation for $x$ we can recycle this value back into the original equation, and see how the answer evolves.

${x}_{2} = 0.0190$, and

${x}_{3} = 0.0191$. This is the so-called (for obvious reasons) $\text{method of successive approximations}$, and most physical scientists use this method. That is get a numerical problem, make an approximation, check the validity of the answer. Is the answer valid? If no, use the approximation to make another approx. How good is the 2nd approx.? And so on..............

Since our approximations have CONVERGED, we have a good value for dissociation, the same as if we solved the initial expression by the quadratic equation.

$\text{%dissociation}$ $=$ "Moles of acid that dissociate"/"Initial moles of acid"xx100% $\cong$ ??%