Benzoic acid reacts with water to form the
benzoate ion by the following reaction
C6H5COOH(aq) + H2O(l)
⇀↽ C6H5COO−(aq) + H3O
+(aq)
The equilibrium constant for this reaction is
6.4 x 10−5
. In a 0.1 M solution of benzoic acid,
what is the concentration of the benzoate ion
at equilibrium?
1. 0.0975 M
2. 2.5 x 10−3 M
3. 0.1 M
4. 6.4 x 10−5 M
5. 8.0 x 10−3 M
1 answer
if benzoic acid ionize partially to x mole of both H30+ and C6H5COO-.At equilibrium, we have 0.1-xM of benzoic acid:6.4*10^-5=x^2/0.1-x since x is relatively small we have:x^2=6.4*10^-5*0.1 x=2.5*10^-3M