Answer:
[tex]3.167\times 10^{-7}[/tex] is the value of [tex]K_a[/tex]for the acid HA.
Explanation:
The pH of the solution = 3.75
The hydrogen ion concentration :
[tex]pH=-log[H^+][/tex]
[tex]3.75=-\log[H^+][/tex]
[tex][H^+]=10^{-3.75}=0.0001778 M[/tex]
Concentration of weak acid = 0.100 mole/L
[tex]HA\rightleftharpoons H^++A^-[/tex]
Initually
0.100 M 0 0
At equilibrium
(0.100-x)M x x
The expression of dissociation constant can be given as:
[tex]K_a=\frac{x\times x}{(0.100-x)}[/tex]
Here the value of x = [tex][H^+]=0.0001778 M[/tex]
[tex]K_a=\frac{0.0001778 M\times 0.0001778 M}{(0.100-0.0001778)M }[/tex]
[tex]K_a=3.167\times 10^{-7}[/tex]
[tex]3.167\times 10^{-7}[/tex] is the value of [tex]K_a[/tex]for the acid HA.
