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Chapter 4: Problem 6

(Integrates with Chapter 2 .) Calculate the concentrations of all ionic species in a \(0.25 M\) solution of histidine at \(\mathrm{pH} 2, \mathrm{pH} 6.4,\) and \(\mathrm{pH} 9.3\).

Short Answer

Expert verified
In conclusion, the concentrations of all ionic species for a 0.25 M solution of histidine are as follows: [H3L] = 0.25 M at pH 2, [H2L^-] = [HL^{2-}] = 0.125 M at pH 6.4, and [L^{3-}] = 0.25 M at pH 9.3.

Step by step solution

01

Understand the Ionization Steps

First, it's important to understand that Histidine is a triprotic acid, meaning it has three replaceable protons. We will denote histidine as H3L. The equilibrium reactions can be given as follows: \n\n1) \(H3L \rightarrow H2L^- + H^+ \)\n2) \(H2L^- \rightarrow HL^{2-} + H^+ \)\n3) \(HL^{2-} \rightarrow L^{3-} + H^+ \)\n\nEach ionization corresponds to a dissociation constant, Ka1, Ka2 and Ka3 respectively.
02

Utilize the Henderson Hasselbalch Equation for pH=2

For the first step of the ionization \(H3L \rightarrow H2L^- + H^+ \), at pH 2 which resides in the first ionization region, we may consider [H2L^-] approximately to be 0. So we can depict the undissociated H3L only. Thus, [H3L] = 0.25 M applies.
03

Calculate the Concentrations for pH=6.4

Now, let's consider the second step of the ionization, \(H2L^- \rightarrow HL^{2-} + H^+ \), at pH 6.4, the pH value is close to the second pKa value, which means that the concentrations of \(H2L^-\) and \(HL^{2-}\) will be approximately equal. So, both are equal to 0.25 M/2.
04

Utilize the Henderson Hasselbalch Equation for pH=9.3

For the third step of ionization, \(HL^{2-} \rightarrow L^{3-} + H^+ \), at pH 9.3, the third step ionize fully since pH is far beyond third pKa. Most of the \(HL^{2-}\) has been converted to \(L^{3-}\). So [\(L^{3-}\)] = 0.25 M in our calculation.

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