To find $-\left[H^{+}\right]=$?
$\quad\quad\quad\quad HA \rightleftharpoons H ^{+}+ A ^{-}$
$t=0 \quad\quad c \quad\quad\quad0\quad \quad 0$
$t=t_1 \quad c-x \quad\quad x \quad\quad x$
For very weak acid $\Rightarrow x\,<<<,1$
$K_a =\frac{\left[H^{+}\right]\left[A^{-}\right]}{[H A]}$
$=\frac{x \times x}{c-x}$
where,$K_a =\text { Dissociation censtent }$
$c =[H A]$
$x =H^{+} \text {concentration }$
As $\quad x\,<<<\,1$ so $c-x \approx c$
$\therefore k_a =\frac{x^2}{c}$
$\Rightarrow x=\sqrt{k_a C}$
|
ક્રમ |
સૂત્ર |
દ્રવ્યતા ગુણાકાર |
|
$1$ |
$PQ$ |
$4.0\times 10^{-20}$ |
|
$2$ |
$PQ_2$ |
$3.2 \times 10^{-14}$ |
|
$3$ |
$PQ_3$ |
$2.7\times 10^{-35}$ |