Answer:
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 ]
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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
