In: Chemistry
Calculate the pH of 0.14 M RbOCl solution
RbOCl(aq) ------------------> Rb^+ (aq) + OCl^- (aq)
0.14M 0.14M
OCl^- (aq) + H2O ----------------> HOCl(aq) + OH^-(aq)
I 0.14 0 0
C -x +x +x
E 0.14-x +x +x
Kb = Kw/Ka
= 1*10^-14/3.5*10^-8
= 2.86*10^-7
Kb = [HOCl][OH^-]/[OCl^-]
2.86*10^-7 = x*x/0.14-x
2.86*10^-7 *(0.14-x) = x^2
x = 0.0002
[OH^-] =x = 0.0002M
POH = -log[OH^-]
= -log0.0002
=3.6989
PH = 14-POH
= 14-3.6989
= 10.3011 >>>>answer