pH of Amphiprotic species & Isoelectric pH

NaHCO3 is amphiprotic species just like NaHSO4, NaHS, Na2HPO4, NaH2PO4. etc.

In such cases the solution pH is dependent on the two ionization constants of the amphiprotic substance and independent of the concentration of the amphiprotic substance in the solution. However, they do not hold pH well and are not effective as buffer solutions. Here the trick is to use the formula,

pH = (pKa1 + pKa2)/2. where Ka1 & Ka2 are dissociation constants of the conjugate acid of amphiprotic species.

For NaHCO3, pH is (6.37 + 10.25)/2 = 8.31

As considered for salts of H3PO4 (pKa1=2.148, pKa2=7.21, pKa3=12.67)-

pH of (H2PO4)- , pH= (pka1 + pka2)/2 hence, For NaH2PO4, pH = 4.67.

For Na2HPO4, pH is 1/2(pKa2+pKa3) = (7.21 + 12.67)/2 = 9.77.

Aminoacids are also an example of amphiprotic species. When an amino acid such as glycine, H2NCH2COOH, is dissolved in water. the carboxylic acid group loses a proton which is gained by the more basic amine group. This produces an ionic structure with opposite charges on both ends, a zwitter-ion(dipolar ion). The zwitter-ion structure of glycine is +H3NCH2COO-. The protonated form of this amphiprotic zwitterion, +H3NCH2COOH, is the glycinium ion. These amino acids are characterised by two pKas, pKa1 and pKa2 for the carboxylic acid and the amine respectively.

Isoelectric pH (pI) is the pH at which aminoacids exists as zwitter-ion. The isoelectronic point(pI) will be halfway between, or the average of, these two pKas, i.e. pI = 1/2 (pKa1 + pKa2). For the simplest amino acid, glycine, pKa1= 2.34 and pKa2 = 9.6, hence pI =(2.34 + 9.6)/2 = 5.97.

Important- (a) Fifteen of 20 amino-acids have isoelectric point near 6. (b) At isoelectric pH, all the molecules have equal positive & negative charges. Hence their is no migration towards any electrode during electrolysis at isolectric pH.

For deivation of pH of amiphiprotic species:

https://www.youtube.com/watch?v=nF6X1PYh45I

Special type- Let us consider the hydrolysis of amphiprotic anion along with cation, e.g., NH4HCO3, NH4HS. In above examples both cations and anions are derived from weak base and weak acids respectively hence, both will undergo hydrolysis in aqueous medium. When these salts are dissolved in water, [H3O+] concentra­tion can be determined as,

[H3O+] = √ka1[kw/kb + ka2]

pH = -log = √ka1[kw/kb + ka2]

Example-

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