Question

A 0.30 solution of a weak acid(HA) has a pH of 3.77. What is the Ka?

Answer 1

Consider dissociation of weak acid HA in water. HA (aq) + H₂O (l)   H₃O ⁺ (aq) + A ^- (aq)  

Dissociation constant of acid HA is, K a = [ H₃O ⁺ ] [ A ^- ] / [ HA ] ----------> (1)

To find out value of Ka , we need to calculate [ H₃O ⁺ ]. We can calculate [ H₃O ⁺ ] from pH value.

We have relation, pH = - log [ H₃O ⁺ ]

[ H₃O ⁺ ] = 10 ^{- pH}

[ H₃O ⁺ ] = 10 ^{- 3.77}

[ H₃O ⁺ ] =  1.70 10 ^-04 M

From reaction, 1 mole of acid HA dissociate & produce 1 mole H₃O ⁺ & 1 mole of A ^-.

[ H₃O ⁺ ] = [ A ^-.] = 1.70 10 ^-04 M

Placing [ H₃O ⁺ ] = [ A ^- ] = 1.70 10 ^-04 M & [ HA ] = 0.30 M in equation 1 , we get

K a = ( 1.70 10 ^-04 ) ( 1.70 10 ^-04 ) / 0.30

Ka = 9.63 10 ^-08

ANSWER : K a of weak acid = 9.63 10 ^-08