From the given graph, when concentration if OCl- is 50 %, then pH is 7.5
pOH = 14 - pH
pOH = 14 - 7.5
pOH = 6.5
When concentration of OCl- reaches 50%, this means half-equivalence point is reached in titration and pKb of OCl- equals to pOH at that point.
pKb = pOH
pKb = 6.5
Kb = 10-pKb
Kb = 10-6.5
Kb = 3.16 x 10-7
initial concentration of OCl- = 0.67 M
ICE table | OCl- (aq) | H2O | HOCl (aq) | OH- | |
Initial conc. | 0.67 M | - | 0 | 0 | |
Change | -x | - | +x | +x | |
Equilibrium conc. | 0.67 M - x | - | +x | +x |
Kb = [HOCl]eq[OH-]eq / [OCl-]eq
3.16 x 10-7 = [(x) * (x)] / (0.67 M - x)
Solving for x, x = 4.6 x 10-4 M
[OH-] = x = 4.6 x 10-4 M
pOH = -log[OH-]
pOH = -log(4.6 x 10-4 M)
pOH = 3.34
pH = 14 - pOH
pH = 14 - 3.34
pH = 10.7
please explain Bleach is a common household cleaner that removes stains, disinfects, and can be used to whiten clot...
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