Calculate the pH of 0.015M NaOCl. No K value is given.
Ka of HClO = 3*10^-8
NaOCl(aq) -------------> Na^+ (aq) + OCl^- (aq)
0.015M ------------------- 0.015M ------- 0.015M
Kb = Kw/Ka
= 1*10^-14/(3*10^-8) = 3.34*10^-7
------------- OCl^- (aq) + H2O(l) -------------> HClO(aq) + OH^- (aq)
I ------------- 0.015 -----------------------------------0 ---------------- 0
C-------------- -x ------------------------------------ +x ----------------- +x
E----------- 0.015-x---------------------------------+x ----------------+x
Kb = [HClO][OH^-]/[OCl^-]
3.34*10^-7 = x*x/(0.015-x)
3.34*10^-7(0.015-x) = x^2
x = 7.06*10^-5
[OH^-] = x = 7.06*10^-5M
POH = -log[OH^-]
= -log(7.06*10^-5)
= 4.15
PH = 14-POH
= 14-4.15
= 9.85 >>>>answer
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