# Question #9a945

Jun 21, 2017

We need $p {K}_{a}$ or ${K}_{a}$ for acetic acid. And under other circumstances I might make you dig for it..........

#### Explanation:

This site tells me that $p {K}_{a} \left(H O A c\right) = 4.76$. Acetic acid is thus a fairly weak acid.......

And now we interrogate the equilibrium......

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

And thus we can write.......

$\frac{\left[{H}_{3} {O}^{+}\right] \left[A c {O}^{-}\right]}{\left[H O A c\right]} = {10}^{- 4.76}$

We know that, initially, $\left[H O A c\right] = 0.540 \cdot m o l \cdot {L}^{-} 1$, and we guess that $x \cdot m o l \cdot {L}^{-} 1$ of acetic acid undergoes protonolysis...

And thus ${10}^{- 4.76} = \frac{x \times x}{0.540 - x} = {x}^{2} / \left(0.540 - x\right)$

This is a quadratic in $x$, which we could solve exactly. Because chemists are lazy folk, however, we ASSUME that $0.540 \text{>>} x$, and thus $0.540 - x \cong 0.540$

And so ${x}_{1} = \sqrt{{10}^{- 4.76} \times 0.540} = 0.00306$

We can recycle this expression, i.e. we use successive approximations, to check on how good it was with respect to the initial assumptions.......

${x}_{2} = \sqrt{{10}^{- 4.76} \times \left(0.540 - 0.00306\right)} = 0.00305$

And this value has converged....so the approximation is good.

Now by definition of our problem $x = \left[{H}_{3} {O}^{+}\right] = 0.00305$, and $p H = 2.52$, and thus $p O H = 11.48$.

And finally, $\left[H {O}^{-}\right] = {10}^{- 11.48} = 3.27 \times {10}^{-} 12 \cdot m o l \cdot {L}^{-} 1$.