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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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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 10^(-10) } for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about calculate the [oh-] in 0.01m aqueous solution of NaOCN {Kb for OCN- is 10^(-10) } covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for calculate the [oh-] in 0.01m aqueous solution of NaOCN {Kb for OCN- is 10^(-10) }.
calculate the [oh-] in 0.01m aqueous solution of NaOCN {Kb for OCN- is 10^(-10) } for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about calculate the [oh-] in 0.01m aqueous solution of NaOCN {Kb for OCN- is 10^(-10) } covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for calculate the [oh-] in 0.01m aqueous solution of NaOCN {Kb for OCN- is 10^(-10) }.
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