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Grade: 12
        
Calculate the [oh-] in the 0.01M aqueous solution of NaOCN {Kb for OCN- = 10^(-10)}
10 months ago

Answers : (2)

BIKRAM
53 Points
							
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)
10 months ago
Saayan Biswas
18 Points
							
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
3 months ago
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