# What is the ratio of lactic acid (Ka = 1.37x10-4) to lactate in a solution with pH =4.59?

Apr 28, 2018

Approximately 1:5

#### Explanation:

If $p H = 4.59$

Then the $\left[{H}_{3} {O}^{+}\right]$ is approximatley $2.57 \times {10}^{-} 5 m o l {\mathrm{dm}}^{-} 3$ as

$p H = - {\log}_{10} \left[{H}_{3} {O}^{+}\right]$

Hence

$\left[{H}_{3} {O}^{+}\right] = {10}^{- p H}$

Because each lactic acid molecule must dissociate to from one lactate ion and one oxonium ion, $\left[{H}_{3} {O}^{+}\right] = \left[l a c t a t e\right]$

If we set up a ${K}_{a}$ expression we can thus find the concentration of the lactic acid:

${K}_{a} = \frac{\left[{H}_{3} {O}^{+}\right] \times \left[l a c t a t e\right]}{\left[L a c t i c .\right]}$

$\left(1.37 \times {10}^{-} 4\right) = {\left(2.57 \times {10}^{-} 5\right)}^{2} / \left(x\right)$

(as it can be assumed that $\left[{H}_{3} {O}^{+}\right] = \left[l a c t a t e\right]$)

Hence

$x = \left[L a c t i c\right] = 4.82 \times {10}^{-} 6$

So,

$\frac{\left[L a c t i c\right]}{\left[L a c t a t e\right]} = \frac{4.82 \times {10}^{-} 6}{2.57 \times {10}^{-} 5} \approx 0.188 \approx 0.2 \approx \left(\frac{1}{5}\right)$

So from this approximation, it seems that the concentration of lactate is almost 5 times higher than that of the lactic acid, so the lactic acid to lactate is (approximately) in a 1:5 ratio.