1. Tardigrade
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  3. Chemistry
  4. pH of mixture of HA and A -buffer is 5 . K b of A -=10-10. Hence [ HA ] /[ A -]will be:

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Q. $pH$ of mixture of $HA$ and $A ^{-}$buffer is $5 . K_{ b }$ of $A ^{-}=10^{-10}$. Hence $[ HA ] /\left[ A ^{-}\right]$will be:

Equilibrium

Solution:

$K_{ b }\left( A ^{-}\right)=10^{-10}$
$p K_{ b }=10$
$p K_{ a }( HA )=14-10=4$
$pH =p K_{ a }+\log \frac{\left[ A ^{-}\right]}{[ HA ]}$
$5=4+\log \frac{\left[ A ^{-}\right]}{[ HA ]}$
$1=\log \frac{\left[ A ^{-}\right]}{[ HA ]}$
$\therefore \frac{\left[ A ^{-}\right]}{[ HA ]}=10$
$\frac{[ HA ]}{\left[ A ^{-}\right]}=\frac{1}{10}=0.1$