In the formula h=(Kh/C)^1/2 where C= concentration so if in the question normality is given 0.01N CH3COOH say for example then will we have to convert it into molarity i.e. 0.03M or we directly substitute 0.01N in place of concentration.Pls give reason for ur answer properly or dont give the answer pls and further solve the question pls.Calculate hydrolysis constant,degree of hydrolysis and ph of 0.01N CH3COONa solution.Ka of CH3COOH=1.8*10^5.pls answer fast and complete?

Explore Courses UPSC Exam UPSC Test Series Free Notes EduRev Infinity

Open App

JEE Exam > JEE Questions > In the formula h=(Kh/C)^1/2 where C= concentr...

Join the Discussion on the App

In the formula h=(Kh/C)^1/2 where C= concentration so if in the question normality is given 0.01N CH3COOH say for example then will we have to convert it into molarity i.e. 0.03M or we directly substitute 0.01N in place of concentration.Pls give reason for ur answer properly or dont give the answer pls and further solve the question pls.Calculate hydrolysis constant,degree of hydrolysis and ph of 0.01N CH3COONa solution.Ka of CH3COOH=1.8*10^5.pls answer fast and complete?

Join for Free

Join for Free

View all answers Start learning for free

Explore Courses for JEE exam

Similar JEE Doubts

Consider a solution of CH3COONH4 which is a salt of weak acid weak base. Theequilibrium involved in the solutions are :If we add these three reactions, then the net reaction isBoth CH3C00- and NH4 get hydrolysed independently and their hydrolysis depends on(i) their initial concentration(ii) the value of Kh which is Since both of the ions were produced from the same salt, their initial concentrations are same . Therefore unless untial the value of Kw/Kaand Kb is same, the degree of hydrolysis of ion cant be same.To explain why we assume that degree of hydrolysis of cation and anion is same, we need to now look at the third reaction i.e., combination of H+ and OH ions. It is obvious that this reaction happens only because one reaction produced H+ ion and the other produced OH ions. We can also note that this reaction causes both the hydrolysis reaction to occur more since their product ions are being consumed. Keep this thing in mind that the equilibrium which has smaller value of equilibrium conxtant is affected more by the common ion effect. For the same reason if for any reason a reaction is made to occur to a greater extent by the consumption of any one of the product ion, the reaction with the smaller value of equilibrium constant tends to get affected more.Therefore we conclude that firstly the hydrolysis of both the ions ocurs more in the presence of each other (due to consumption of the product ions) than in each other is absence. Secondly the hydrolysis of the ion which occurs to a lesser extent (due to smaller value of Kh) is affected more than the one whose Kh is greater. Hence we can see that the degree of hydrolysis of both the ions would be close to each other when they are getting hydrolysed in the presence of each other.Q.In a solution of NaHCO3 , the amphiprotic anion can under ionization to form H+ ion and hydrolysis to from OH ion.To calculat PH, suitable approximation is

View answer

The rate of chemical reaction at a particular temperature is proportional to the product of the molar concentration of reactants with each concentration term raised to the power equal to the number of molecules of the respective reactant taking part in the reaction. aA + bB products, rate of reaction a [Aka [Bib = k [Aka [Bib, where k is the rate constant of the reaction.EQUILIBRIUM CONSTANT (k) For a general reaction aA + bB = cC + dD, forward rate rf = kf [A]a [B]b, backward rate rb = kb [Cicpyi ,concentrations of reactants products at equilibrium are related bywhere all theconcentrations are expressed in mole/liter.In the expression of equilibrium constant those components are kept whose concentration changes with time.If equilibrium is estabilished by taking all the components in the reaction then to predict the direction of reactions we calculate the reaction quotient (Q).The values of expression at any time during reaction is called reaction quotient if Q Kc reaction proceed in backward direction until equilibrium in reached if Q Kc reaction will proceed in forward direction until equilibrium is established if Q = KC Reaction is at equilibriumq.Above homogeneous reaction is carried out in a 2 litre container at a particular temperature by taking 1 mole each of A, B, C and D respectively. If Kc for the reaction is then equilibrium concentration of C is

View answer

The rate of chemical reaction at a particular temperature is proportional to the product of the molar concentration of reactants with each concentration term raised to the power equal to the number of molecules of the respective reactant taking part in the reaction. aA + bB products, rate of reaction a [Aka [Bib = k [Aka [Bib, where k is the rate constant of the reaction.EQUILIBRIUM CONSTANT (k) For a general reaction aA + bB = cC + dD, forward rate rf = kf [A]a [B]b, backward rate rb = kb [Cicpyi ,concentrations of reactants products at equilibrium are related bywhere all theconcentrations are expressed in mole/liter.In the expression of equilibrium constant those components are kept whose concentration changes with time.If equilibrium is estabilished by taking all the components in the reaction then to predict the direction of reactions we calculate the reaction quotient (Q).The values of expression at any time during reaction is called reaction quotient if Q Kc reaction proceed in backward direction until equilibrium in reached if Q Kc reaction will proceed in forward direction until equilibrium is established if Q = KC Reaction is at equilibriumQ.Initial concentration of A is twice the initial concentration of `B. At equilibrium concentration of A and a are same then equilibrium constant for the reaction is

View answer

An optically active compound A upon acid catalysed hydrolysis yield two optically active compound B and C by pseudo first order kinetics. The observed rotation of the mixture after 20 min was 5 while after completion of the reaction it was 20. If optical rotation per mole of A, B C are 60, 40 80. Calculate half life of the reaction. Correct answer is '20'. Can you explain this answer?

View answer

In a second order reaction, the initial concentration of the reactants is 0.1 mol/L. The reaction is found to be 20% complete in 40 minutes. Calculate (i) the rate constant (ii) half-life period (iii) time required to complete 75% of the reaction.?

View answer

Top Courses for JEEView all

Physics for JEE Main & Advanced

Mock Tests for JEE Main and Advanced 2025

Chemistry for JEE Main & Advanced

Mathematics (Maths) for JEE Main & Advanced

Chapter-wise Tests for JEE Main & Advanced

Top Courses for JEE

View all

Share with a Friend

Answer this doubt

Question Description
In the formula h=(Kh/C)^1/2 where C= concentration so if in the question normality is given 0.01N CH3COOH say for example then will we have to convert it into molarity i.e. 0.03M or we directly substitute 0.01N in place of concentration.Pls give reason for ur answer properly or dont give the answer pls and further solve the question pls.Calculate hydrolysis constant,degree of hydrolysis and ph of 0.01N CH3COONa solution.Ka of CH3COOH=1.8*10^5.pls answer fast and complete? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about In the formula h=(Kh/C)^1/2 where C= concentration so if in the question normality is given 0.01N CH3COOH say for example then will we have to convert it into molarity i.e. 0.03M or we directly substitute 0.01N in place of concentration.Pls give reason for ur answer properly or dont give the answer pls and further solve the question pls.Calculate hydrolysis constant,degree of hydrolysis and ph of 0.01N CH3COONa solution.Ka of CH3COOH=1.8*10^5.pls answer fast and complete? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In the formula h=(Kh/C)^1/2 where C= concentration so if in the question normality is given 0.01N CH3COOH say for example then will we have to convert it into molarity i.e. 0.03M or we directly substitute 0.01N in place of concentration.Pls give reason for ur answer properly or dont give the answer pls and further solve the question pls.Calculate hydrolysis constant,degree of hydrolysis and ph of 0.01N CH3COONa solution.Ka of CH3COOH=1.8*10^5.pls answer fast and complete?.

Solutions for In the formula h=(Kh/C)^1/2 where C= concentration so if in the question normality is given 0.01N CH3COOH say for example then will we have to convert it into molarity i.e. 0.03M or we directly substitute 0.01N in place of concentration.Pls give reason for ur answer properly or dont give the answer pls and further solve the question pls.Calculate hydrolysis constant,degree of hydrolysis and ph of 0.01N CH3COONa solution.Ka of CH3COOH=1.8*10^5.pls answer fast and complete? in English & in Hindi are available as part of our courses for JEE. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.

Here you can find the meaning of In the formula h=(Kh/C)^1/2 where C= concentration so if in the question normality is given 0.01N CH3COOH say for example then will we have to convert it into molarity i.e. 0.03M or we directly substitute 0.01N in place of concentration.Pls give reason for ur answer properly or dont give the answer pls and further solve the question pls.Calculate hydrolysis constant,degree of hydrolysis and ph of 0.01N CH3COONa solution.Ka of CH3COOH=1.8*10^5.pls answer fast and complete? defined & explained in the simplest way possible. Besides giving the explanation of In the formula h=(Kh/C)^1/2 where C= concentration so if in the question normality is given 0.01N CH3COOH say for example then will we have to convert it into molarity i.e. 0.03M or we directly substitute 0.01N in place of concentration.Pls give reason for ur answer properly or dont give the answer pls and further solve the question pls.Calculate hydrolysis constant,degree of hydrolysis and ph of 0.01N CH3COONa solution.Ka of CH3COOH=1.8*10^5.pls answer fast and complete?, a detailed solution for In the formula h=(Kh/C)^1/2 where C= concentration so if in the question normality is given 0.01N CH3COOH say for example then will we have to convert it into molarity i.e. 0.03M or we directly substitute 0.01N in place of concentration.Pls give reason for ur answer properly or dont give the answer pls and further solve the question pls.Calculate hydrolysis constant,degree of hydrolysis and ph of 0.01N CH3COONa solution.Ka of CH3COOH=1.8*10^5.pls answer fast and complete? has been provided alongside types of In the formula h=(Kh/C)^1/2 where C= concentration so if in the question normality is given 0.01N CH3COOH say for example then will we have to convert it into molarity i.e. 0.03M or we directly substitute 0.01N in place of concentration.Pls give reason for ur answer properly or dont give the answer pls and further solve the question pls.Calculate hydrolysis constant,degree of hydrolysis and ph of 0.01N CH3COONa solution.Ka of CH3COOH=1.8*10^5.pls answer fast and complete? theory, EduRev gives you an ample number of questions to practice In the formula h=(Kh/C)^1/2 where C= concentration so if in the question normality is given 0.01N CH3COOH say for example then will we have to convert it into molarity i.e. 0.03M or we directly substitute 0.01N in place of concentration.Pls give reason for ur answer properly or dont give the answer pls and further solve the question pls.Calculate hydrolysis constant,degree of hydrolysis and ph of 0.01N CH3COONa solution.Ka of CH3COOH=1.8*10^5.pls answer fast and complete? tests, examples and also practice JEE tests.

Explore Courses for JEE exam