Respuesta :

W0lf93

Answer:  
NaZ + H2O <----> HZ + OH- 
 [OH-] = [HZ] 
 [NaZ] = 0.15 M 
 pH = 10.7, so the pOH = 14-10.7 = 3.3; [OH-] = 10^-3.3 = 5.01E-4 
 kb = [OH-][HZ]/[NaZ] = (5.01E-4)^2/0.15 = 1.67E-6 
 ka = kw/kb = 1E-14/1.67E-6 = 5.99E-9 = 6.0E-9

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