Kb Oh

Kb=[OH-][HOCL]/[OCl-] (remember all these are equilllibrium constrations of each specie involved) when we look on Kb table we are unlucky to get the value of (Kb), but don't worry ...

[oh-] = 10^-3.266 = 5.42e-4 tea + h2o <====> teah+ + oh-kb = [oh-][teah+] / [tea] [oh-] = [teah+] = 5.42e-4 kb = (5.42e-4)(5.42e-4) / 0.474 = 6.20e-7

Kb = [OH⁻]² / [CN⁻] => [CN⁻] = [OH⁻]² / Kb = 10^(-3.8) / 10^(-4.8) = 10M The number of moles of dissolved sodium cyanide per volume is equal to the cyanide ion concentration ...

Kb = [OH-][PHNH3+]/[PHNH2] = 7.4 *10-10 at eq [OH-]=[PHNH3+]=x so x2/.4 = /7.4*10-10 ==> x = 1.72*10-5M [OH-] pOH = -log[OH-] = 4.76 pOH + pH = 14 ==> pH = 9.24

[BO] ---> [B][O] so Kb= [OH-]^2 / ([Base]-[OH-]) But we now the concentration of the solution and the OH- so adding in these values we get... Kb= (6.60*10^-5)^2 / (.10 - 6.60*10^-5)