In: Chemistry
Determine the Ph
0.120 M NaClO and 4.50×10−2 M KI
Express your answer using two decimal places.
KI is a slat of strong HI and strong base . so it will not produce H+ and OH- , so no effect on PH
NaClO----------> Na+ + OCl-
OCl- + H2O -------> HoCl + OH-
I 0.12 0 0
C -x +x +x
E 0.12-x +x +x
Kb = [HOCl][OH-]/[OCl-]
Kb = Kw/Ka
= 1*10-14/3*10-8
= 3.3*10-7
Kb = [HOCl][OH-]/[OCl-]
3.3*10-7 = x*x/0.12-x
3.3*10-7*(0.12-x) = x2
x = 0.000198
[OH-] = x = 0.000198M
POH = -log[OH-]
= -log0.000198
= 3.703
PH = 14-POH
= 14-3.703
= 10.297 >>>> answer