Question

The dissociation constant of weak acid is 1×10-4 in order to prepare a buffer solution with ph =5 the salt acid ratio

Answer 1

Answer:The salt acid ratio is 10:1

Explanation:

Dissociation constant of weak acid = $1\times 10^{-4}$

the pH of weak acid = 5

$HA+H_2O\rightleftharpoons H_3O^++A^-$

$K_a= \frac{[H_3O^+][A^-]}{[HA]}$

$[H_3O^+]=\frac{K_a\times [HA]}{[A^-]}$

$pH=5=-\log[H_3O^+]=-\log[\frac{K_a\times [HA]}{[A^-]}]=-\log{K_a}-\log\frac{[HA]}{[A^-]}$

$5=-\log(1\times 10^{-4})+\log\frac{[A^-]}{[HA]}$

$\frac{[A^-]}{[HA]}=10:1$

The salt acid ratio is 10:1

Answer 2

Answer : The salt acid ratio is 10 : 1

Solution : Given,

Dissociation constant of weak acid = $1\times 10^{-4}$

pH of weak acid = 5

The equilibrium reaction for dissociation of weak acid will be,

$HA+H_2O\rightleftharpoons H_3O^++A^-$

The expression for dissociation constant will be,

$K_a=\frac{[H_3O^+][A^-]}{[HA]}$

$[H_3O^+]=\frac{K_a\times [HA]}{[A^-]}$

$pH=-\log[H_3O^+]$

Now put the value of $[H_3O^+]$ in pH expression, we get

$pH=-\log[\frac{K_a\times [HA]}{[A^-]}]=-\log{K_a}-\log\frac{[HA]}{[A^-]}$

Now put all the given values in this expression, we get the salt acid ratio.

$5=-\log(1\times 10^{-4})-\log\frac{[HA]}{[A^-]}$

$\frac{[HA]}{[A^-]}=1:10$

or,

$\frac{[A^-]}{[HA]}=10:1$

Therefore, the salt acid ratio is 10:1