Kh = (OH^-)(HClO)/(ClO^-).
Let x = (HClO), then x = (OH^-)
and (ClO^-) = 7.68 x 10^-4 -x
Solve for x.
What is the molarity of OH- in a 7.68×10-4 M NaClO solution that hydrolyzes according to the equation.
ClO-(aq) + H2O(l) = OH-(aq) + HClO(aq)
Constants: Kh=2.86×10-7
Can someone help me figure this one out?
1 answer