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................. HClO(ac) + H2O(l) <----> ClO- (ac) + H3O+ (ac)
inicial: ..........a........................................0...................0
cambio: .......-X......................................+X................+X
final: ...........a-X.......................................X..................X
Ka = ( [ClO-] . [H3O+] ) ÷ [HClO]
Ka = x^2 ÷ (a - X)
donde:
Ka = 0.011
X = [H3O+] = 10^-pH = 10^-2.1 = 7.94×10^-3 M
Sustituyendo y despejando "a" del equilibrio
a = 0.0137M
m HClO = 2L × ( 0.0137 mol / L ) ---> 0.027 mol