A 0.199 M weak acid solution has a pH of 3.99. Find Ka for the acid.

Ka = ( About ) 5 x 10^ - 8

Explanation:

Let us first identify the dissociation equation for this weak acid,

HA ⇌ ( H+ ) + A¯

Knowing this, we can tell what the equilibrium expression is, respectively,

Ka = ( [ H+ ] [ A¯ ] ) / [ HA ]

Now let us use the given pH 3.99 to calculate [ H+ ], knowing that pH = −log [H+],

3.99 = - log[H+],

[H+] = 10 ^ - 3.99,

[H+] = ( About ) 1 * 10^-4 M

Substitute known values into the equilibrium expression,

Ka = [( 1 x 10^ - 4 ) ( 1 x 10^ ¯4 )] / 0.199,

Ka = ( About ) 5 x 10^ - 8