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Calculate the [oh-] in the 0.01M aqueous solution of NaOCN {Kb for OCN- = 10^(-10)}
NaOCN is a salt of weak acid and strong base .So POH=(1/2)×Pkw + (1/2)×Pkb + (1/2)×logC POH= 7 + 0.5× 10 + 0.5×2 = 7 +5+1=13[OH-] =10−13(M)
As NaOCN is salt of weak acid and strong base NaOCN will dissociate and the OCN- will hydrolyse acting as a base(proton acceptor) while water will act as acid(proton donor)NaOCN==> Na+. + OCN- OCN- +H2O HOCN + OH-C(1-a). Ca . Ca. a=degree of dissociation Ca=[OH-] = √(KbC) =10^-6. So pOH=6
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