J.R. S. answered 10/01/22
Ph.D. University Professor with 10+ years Tutoring Experience
Set up an ICE table:
A(aq) <===> 2B(aq)
1.50.................0...........Initial
-x...................+2x.........Change
1.50-x.............2x..........Equilibrium
K = [B]2[A]
3.04x10-6 = (2x)2(1.50-x) and because K is a low value, we can assume x is small and ignore it
3.04x10-6 = (4x2 )(1.50)
4x2 =2.03x10-6
x2= 5.07x10-7
x = 7.12x10-4 (which is small compared to 1.50 so the above assumption was valid)
[B] at equilibrium = 2x = (2)(7.12x10-4) = 1.42x10-3 M