Question

What is the ph of 0.01 M glycine solution for glycine ka1=4.5×10^-3 and ka2=1.7×10^-100 at 298k

Answer 1

 the ph of 0.01 M glycine solution for glycine ka1=4.5×10^-3 and ka2=1.7×10^-100 at 298k is 3

Answer 2

The pH of 0.01 M glycine solution for glycine with $K_{a1} = 4.5\times10^{-3}$ and $K_{a2} = 1.7\times10^{-10}$ is 7.06.

Given:

$K_{a1} = 4.5\times10^{-3}$

$K_{a2} = 1.7\times10^{-10}$

To find:

pH = ?

Formula to be used:

pH = $-log[H^+]$

Calculation:

To find $H^+$ concentration, the value of K is required.

K = $K_{a1}+K_{a2}$

K = $(4.5\times10^{-3})+ (1.7\times10^{-10})$

K = $7.65\times10^{-13}$

Value of $H^+$ concentration

$H^+$ = K× C

$H^+$ =( $7.65\times10^{-13}$) × 0.01

$H^+$ = 0.87 mol $L^{-1}$

pH =  $-log[H^+]$

pH = $-log[0.87\times10^{-7}]$

pH = 7.06

Conclusion:

The pH of 0.01 M glycine solution for glycine with $K_{a1} = 4.5\times10^{-3}$ and $K_{a2} = 1.7\times10^{-10}$ is 7.06.

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