Question

I am confusing the rules of pka and pH. Please explain fully I am so confused!!

6. Diabetic acidosis is caused by an inability to metabolize glucose, forcing metabolism of fatty acids from adipose tissue for energy. The condition is a complication of type I diabetes (juvenile diabetes) and results from failure to secrete insulin. Diabetic acidosis causes an increase in blood levels of the following compounds: Оно "ОН B-hydroxy butyric acid (BHB) pKa 4.7 Он acetoacetic acid (ACA) pKa 3.6 Normal blood has a pH of 7.4 while in severe acidosis patients can have blood pH levels near 6.0. a) Which solution is the most acidic? А. 5 mM ACА В. З mM BHВ C. 3 mM ACA Explain your reasoning. b) In a solution of BHB in which pH 6.0, what is the ratio of [HA] to [A-]?

Answer 1

a) The pKa of a weak acid is the negative logarithmic measure of the Ka of the acid where Ka is the equilibrium constant for the dissociation of the weak acid. Since strong acids dissociate completely without establishing an equilbrium, ideally they have a Ka of 1 while weak acids have Ka less than 1 always.

For a weak acid HA, the Ka for the given equilibrium is as follows.

$HA \rightleftharpoons H^+ + A^-$

$K_a = \frac{[H^+][A^-]}{[HA]}$

$pK_a = -log\;K_a$

The stronger an acid, the greater will be its dissociation. So, the value of Ka will grow closer to unity. As Ka increases, pKa decreases since it is the negative of the logarithm of Ka. Thus, stronger acids will have lower pKa values while weaker acids will have greater pKa values.

Now, it can be ascertained that among BHB and ACA, ACA is the stronger acid. So, a solution of ACA will be more acidic than a solution of BHB of the same concentration. Also, since the pH of a weak acid is proportional to its Ka and concentration, 5mM of ACA will be more acidic than 3 mM of ACA. So, the most acidic solution among those given will be 5 mM of ACA.

b) At a particular pH, weak acids dissociate to a certain extent and form their conjugates. The ratio of the conjugate concentration to the undissociated acid's concentration is given by the Henderson-Hasselbach equation, which is as follows.

$pH = pK_a + log\frac{[A^-]}{[HA]}$

For a solution of BHB at pH = 6.0, it will be

$6.0 = 4.7 + log\frac{[A^-]}{[HA]}$

$log\frac{[A^-]}{[HA]} = 1.3 \Rightarrow \frac{[A^-]}{[HA]} = 10^{1.3}$

This gives the ratio of A- to HA as 19.9526 : 1.