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

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

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Answer 1 · 2018-04-28T13:34:56

Approximately 1:5

Explanation:

If $p H = 4.59$ $pH=4.59$

Then the $[H_{3} O^{+}]$ $[H_3O^(+)]$ is approximatley $2.57 \times 10^{- 5} m o l d m^{- 3}$ $2.57 times 10^-5 moldm^-3$ as

$p H = - {log}_{10} [H_{3} O^{+}]$ $pH=-log_10[H_3O^(+)]$

Hence

$[H_{3} O^{+}] = 10^{- p H}$ $[H_3O^(+)]=10^(-pH)$

Because each lactic acid molecule must dissociate to from one lactate ion and one oxonium ion, $[H_{3} O^{+}] = [l a c t a t e]$ $[H_3O^(+)]=[lactate]$

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

$K_{a} = \frac{[H_{3} O^{+}] \times [l a c t a t e]}{[L a c t i c .]}$ $K_a=([H_3O^(+)] times [lactate])/([Lactic.])$

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

(as it can be assumed that $[H_{3} O^{+}] = [l a c t a t e]$ $[H_3O^(+)]=[lactate]$ )

Hence

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

So,

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

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.

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

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