# How do you determine Ka?

Feb 22, 2017

$\text{How else but by experimental measurement?}$

#### Explanation:

Of course you must have some means to measure the equilibrium. In aqueous solution, we address the equilibrium:

$H A \left(a q\right) + {H}_{2} O \left(l\right) \rightarrow {A}^{-} + {H}_{3} {O}^{+}$

Given a $\text{glass electrode}$, an electrode sensitive to ${H}_{3} {O}^{+}$, we can measure the concentration of ${H}_{3} {O}^{+}$, which given the stoichiometry of the reaction is equal to the concentration of ${A}^{-}$, and the starting concentration of $H A$ should also be known. And thus:

${K}_{a} = \frac{\left[{A}^{-}\right] \left[{H}_{3} {O}^{+}\right]}{\left[H A\right]}$

And so if (say) the glass electrode gives a value for $\left[{H}_{3} {O}^{+}\right]$, say $\left[{H}_{3} {O}^{+}\right] = x$.

And thus: ${K}_{a} = \frac{{x}^{2}}{\left[H A\right] - x}$.

This is quadratic in $x$, that may be solved exactly, given that we know the starting value of $\left[H A\right]$. Typically, simple approximations are used so we do not have to pfaff about with quadratics, i.e.

${x}_{1} \cong \sqrt{{K}_{a} \left[H A\right]}$

Successive approximations (${x}_{1} , {x}_{2} , {x}_{3}$) rapidly give a value for $x$ that is identical to the value that would be obtained by the quadratic equation.