Question

Dissociation constant for an acid HA is 1.6x10-5.calculate its H3O in 0.01 M solution

Answer 1

Answer:

Answer

4.0×10−4 mol L−1

Solution

the dissociation of the acid HA may be represented as :

HA(aq)⇔H+(aq)+A−(aq)

Ka=[H+(aq)][A−(aq)][HA(aq)]=[H+(aq)]2[HA(aq)]

Ka=1.6×10−5,[HA]=0.01M

1.6×10−5=[H+(aq)]20.01 or [H+(aq)]=1.6×10−5×0.01−−−−−−−−−−−−−−√=4.0×10−4mol L−1

Explanation: