Lactic acid is a weak acid with the formula C 3 H 6 O 3 that can be found in sou

Question

Lactic acid is a weak acid with the formula C3H6O3that can be found in sour milk products like yogurt, and contributes to the natural curdling of milk (a process you may remember from lab 3). The pKaof lactic acid is 3.85 at 298 K. The lactate content of blood is sometimes used to measure the acid-base homeostasis within the body. The acceptable concentration range measured arterially is considered to be 0.5 to 1.6 mM.

What would be the pH of a 1.6 mM solution of Lactate? For the purpose of this question, assum percent ionization is less than 5%. Then test you rassumption by calculating the percent of ionization of lactate in the 1.6 mM solution.

Explanation / Answer

The pKa of lactic acid = 3.85; therefore, Ka = antilog(-pKa) = antilog (-3.85) = 1.4125*10-4.

We know that Ka(lactic acid)*Kb(lactate) = Kw where Kw = 1.0*10-14. Therefore.

Kb(lactate) = Kw/Ka = (1.0*10-14)/(1.4125*10-4) = 9.9115*10-11.

Consider the ionization of lactate (denote as B) as below.

B (aq) + H2O (l) <=======> BH+ (aq) + OH- (aq)

Since OH- is a base, we work with Kb.

Kb = [BH+][OH-]/[B]

====> 9.9115*10-11 = (x).(x)/(1.6 – x)

We assume the extent of ionization, x is less than 5%,; hence, (1.6 – x) mM ? 1.6 mM. Solve for x.

9.9115*10-11 = x2/(1.6)

====> x2 = 9.9115*10-11*1.6 = 1.58584*10-10

====> x = 1.2593*10-5

Therefore, [OH-] = 1.2593*10-5 mM = (1.2593*10-5 mM)*(1 M/1000 mM) = 1.2593*10-8 M and pOH = -log [OH-] = -log (1.2593*10-8) = 7.8999.

We know that pH + pOH = 14; therefore, pOH = 14 – pH = 14 – 7.8999 = 6.1001 ? 6.1 (ans).

We have [BH+] = [OH-] = 1.2593*10-5 mM and [B] = 1.6 mM; therefore, percent ionization = [BH+]/[B]*100 = (1.2593*10-5 mM)/(1.6 mM)*100 = 7.870625*10-4 ? 7.87*10-4%. The extent of ionization is indeed much less than 5% (ans).

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