Isoelectric point of alanine is 6-0 while ionization constant of COOH of alanine is 5-0 - 10-3- Find the pH at which concentrations of the given species are same-log 5-0-7-

Given, K_a1=5.0 × 10^ 3 So, pK_a1= log (K_a1)=2.3 Isoelectric point, pI= pK_a1+pK_a2/2 ⇒ 6.0 = 2.3+pK_a2/2 ⇒ pK_a2 = 9.7 When concentration of both species are ...

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Standard XII

Chemistry

Stability of Carbocation

Isoelectric p...

Question

Isoelectric point of alanine is $6.0$ while ionization constant of $C O O H$ of alanine is $5.0 \times 10^{- 3}$ . Find the $p H$ at which concentrations of the given species are same.

[log 5 = 0.7]

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Solution

Given, $K_{a 1} = 5.0 \times 10^{- 3}$
So, $p K_{a 1} = - l o g (K_{a 1}) = 2.3$
Isoelectric point, $p I = \frac{p K_{a 1} + p K_{a 2}}{2} \Rightarrow 6.0 = \frac{2.3 + p K_{a 2}}{2} \Rightarrow p K_{a 2} = 9.7$
When concentration of both species are same then $p K_{a 2}$ will be equal to pH.

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Isoelectric point of alanine is 6.0 while ionization constant of COOH of alanine is 5.0×10−3. Find the pH at which concentrations of the given species are same. [log 5 = 0.7]

Given, Ka1=5.0×10−3 So, pKa1=−log (Ka1)=2.3 Isoelectric point, pI=pKa1+pKa22⇒6.0=2.3+pKa22⇒pKa2=9.7 When concentration of both species are same then pKa2 will be equal to pH.

Isoelectric point of alanine is $6.0$ while ionization constant of $C O O H$ of alanine is $5.0 \times 10^{- 3}$ . Find the $p H$ at which concentrations of the given species are same.

[log 5 = 0.7]

Given, $K_{a 1} = 5.0 \times 10^{- 3}$
So, $p K_{a 1} = - l o g (K_{a 1}) = 2.3$
Isoelectric point, $p I = \frac{p K_{a 1} + p K_{a 2}}{2} \Rightarrow 6.0 = \frac{2.3 + p K_{a 2}}{2} \Rightarrow p K_{a 2} = 9.7$
When concentration of both species are same then $p K_{a 2}$ will be equal to pH.