The equilibrium constants for amino acids are given in terms of succesive ionization constants of the protonated form.For example, equilibrium constats for Glycine `(NH_2CH_2COOH)` are `K_(a1)=5xx10^(-3) M and K_(a2)=2xx10^(-10)M`.What will be the pH at the Isoelectric point for this amino acid and pH of 0.02 M protonated Glycine in pure water respectively ?[Take :log 2=0.30]

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Correct Answer - 62
Isoelectric point pH =`1/2[pKa_(1)+pKa_(2)]=1/2[2.3+9.7]=6`
`[H^+]=sqrt(5xx10^(-3)Mxx0.2)=sqrt(10xx10^(-5))=10^(-2)`
pH=-log `[H^+]=2`

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