Question

) at pH 8.267 The acid What fraction of piperazine (perhydro-1,4-diazine) is in each of its three forms (H.A, HA, A dissociation constant values for piperazine are Kl = 4.65 x 10 and K2 = 1.86 x 10-10 LA - =

Answer 1

We have equation, $\alpha$_H2A =  [H⁺] ² / [H⁺] ² + K ₁ [H⁺] + K ₁ K ₂

We have given K ₁ = 4.65 10 ^-06  , K ₂ = 1.86 10 ^-10

We have, pH = - log [H⁺] .

Therefore, [H⁺] = 10 ^{- pH} = 10 ^{- 8.26} = 5.495 10 ^-09 M

Putting above values in equation, we get

$\alpha$_H2A = (5.495 10 ^-09 ) ² / (5.495 10 ^-09) ² + ( 4.65 10 ^-06   5.495 10 ^-09 ) + ( 4.65 10 ^-06 )( 1.86 10 ^-10 )

$\alpha$_H2A = 3.020 10 ^-17 / 3.020 10 ^-17 + 2.555 10 ^-14 + 8.649 10 ^-16

$\alpha$_H2A = 3.020 10 ^-17 / 2.644 10 ^-14

$\alpha$_H2A = 0.00114

We have, $\alpha$_{HA ^-} =  ( K ₁ [H⁺] ) / [H⁺] ² + K ₁ [H⁺] + K ₁ K ₂

$\alpha$_{HA ^-} = ( 4.65 10 ^-06 ) (5.495 10 ^-09 ) / (5.495 10 ^-09) ² + ( 4.65 10 ^-06   5.495 10 ^-09 ) + ( 4.65 10 ^-06 )( 1.86 10 ^-10 )

$\alpha$_{HA ^-} = 2.555 10 ^-14 /  3.020 10 ^-17​​​​​​​ + 2.555 10 ^-14 + 8.649 10 ^-16

$\alpha$_{HA ^-} = 2.555 10 ^-14 / 2.644 10 ^-14

$\alpha$_{HA ^-} = 0.966

We have, $\alpha$_{A ^2-}   = K ₁ K ₂ / [H⁺] ² + K ₁ [H⁺] + K ₁ K ₂

$\alpha$_{A ^2-}   = ( 4.65 10 ^-06 )( 1.86 10 ^-10 ) / (5.495 10 ^-09) ² + ( 4.65 10 ^-06   5.495 10 ^-09 ) + ( 4.65 10 ^-06 )( 1.86 10 ^-10 )

$\alpha$_{A ^2-}   = 8.649 10 ^-16 /  3.020 10 ^-17​​​​​​​ + 2.555 10 ^-14 + 8.649 10 ^-16

$\alpha$_{A ^2-}   = 8.649 10 ^-16 / 2.644 10 ^-14

$\alpha$_{A ^2-}   = 0.0327