Class 11 Exam > Class 11 Questions > calculate the [oh-] in 0.01m aqueous solution...
Start Learning for Free
calculate the [oh-] in 0.01m aqueous solution of NaOCN {Kb for OCN- is 10^(-10) }
Most Upvoted Answer
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
Attention Class 11 Students!
To make sure you are not studying endlessly, EduRev has designed Class 11 study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in Class 11.
|
Explore Courses for Class 11 exam
|
|
Similar Class 11 Doubts
Top Courses for Class 11View all
calculate the [oh-] in 0.01m aqueous solution of NaOCN {Kb for OCN- is 10^(-10) }
Question Description
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) }.
Solutions for calculate the [oh-] in 0.01m aqueous solution of NaOCN {Kb for OCN- is 10^(-10) } in English & in Hindi are available as part of our courses for Class 11.
Download more important topics, notes, lectures and mock test series for Class 11 Exam by signing up for free.
Here you can find the meaning of calculate the [oh-] in 0.01m aqueous solution of NaOCN {Kb for OCN- is 10^(-10) } defined & explained in the simplest way possible. Besides giving the explanation of
calculate the [oh-] in 0.01m aqueous solution of NaOCN {Kb for OCN- is 10^(-10) }, a detailed solution for calculate the [oh-] in 0.01m aqueous solution of NaOCN {Kb for OCN- is 10^(-10) } has been provided alongside types of calculate the [oh-] in 0.01m aqueous solution of NaOCN {Kb for OCN- is 10^(-10) } theory, EduRev gives you an
ample number of questions to practice calculate the [oh-] in 0.01m aqueous solution of NaOCN {Kb for OCN- is 10^(-10) } tests, examples and also practice Class 11 tests.
|
|
Explore Courses for Class 11 exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup