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calculate the [oh-] in 0.01m aqueous solution of NaOCN {Kb for OCN- is 10^(-10) }
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calculate the [oh-] in 0.01m aqueous solution of NaOCN {Kb for OCN- is...
Calculation of [OH-] in 0.01M Aqueous Solution of NaOCN


Understanding the Problem


In this problem, we are given a 0.01M aqueous solution of NaOCN. Our task is to calculate the concentration of hydroxide ions ([OH-]) in this solution. We are also given the value of Kb for OCN- which is 10^(-10).


Writing the Chemical Equation


NaOCN is a salt of a weak acid (HCN) and a strong base (NaOH). Therefore, it will undergo hydrolysis in water to form HCN and OH- ions. The chemical equation for this reaction is:


NaOCN + H2O ↔ HCN + NaOH

Writing the Equilibrium Expression


Using the above chemical equation, we can write the equilibrium expression for this reaction as follows:


Kb = [HCN][OH-] / [OCN-]

Using the Given Information to Find [OH-]


As we are given the value of Kb for OCN-, we can use this value to find the concentration of OH- ions in the solution.


Kb = 10^(-10) = [HCN][OH-] / [OCN-]

[HCN] = [NaOH] = 0.01 M (as NaOH is a strong base and is completely dissociated)

[OCN-] = [NaOCN] = 0.01 M (as NaOCN is completely dissociated)

Substituting these values in the equilibrium expression, we get:

10^(-10) = [HCN][OH-] / [OCN-]

10^(-10) = (0.01)([OH-]) / 0.01

[OH-] = 10^(-8)

Therefore, the concentration of hydroxide ions in the given solution is 10^(-8) M.

Conclusion


Using the given information and the concept of hydrolysis, we were able to calculate the concentration of hydroxide ions in the given solution. The value of Kb for OCN- helped us to relate the concentrations of various species in the equilibrium expression and solve for the unknown concentration.
Community Answer
calculate the [oh-] in 0.01m aqueous solution of NaOCN {Kb for OCN- is...
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
So [OH-]=10^-6
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