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In the salt hydrolysis from strong foot and you will weakened acidic, we need to obtain a love ranging from K

28 juin In the salt hydrolysis from strong foot and you will weakened acidic, we need to obtain a love ranging from K

Matter 5. The concentration of hydronium ion inside acid buffer provider hinges on the new proportion of intensity of the weak acid into the concentration of the conjugate foot contained in the answer. i.elizabeth.,

dos. The brand new weak acid was dissociated only to a little the amount. Moreover due to popular ion perception, this new dissociation is further stored so because of this this new harmony concentration of brand new acid is nearly comparable to the initial concentration of the latest unionised acid. Likewise new intensity of the fresh new conjugate legs is nearly comparable to the original concentration of the additional sodium.

step three. [Acid] and you may [Salt] represent the initial intensity of new acidic and you can salt, respectively accustomed ready yourself new barrier provider.

Question 6. Explain about the hydrolysis of salt of strong acid and a strong base with a suitable example. Answer: 1. Let us consider the neutralisation reaction between NaOH and HNO3 to give NaNO3 and water. NaOH(aq) + HNO3(aq) > NaNO3(aq) + H2O(1)

3. Water dissociates to a small extent as H2O(1) H + (aq) + OH – (aq) Since [H + ] = [OH – ], water is neutral.

cuatro. NO3 ion is the conjugate base of strong acid HNO3 and hence it has no tendency to react withH + ,

Derive Henderson – Hasselbalch picture Answer: step 1

5. Similarly Na ‘s the conjugate acid of your good ft NaOH and has now no tendency to work with OH

six. It means that there surely is zero hydrolysis. In these instances [H + ] (OH – ), pH are maintained so there fore the answer try natural.

Question 7. Explain about the hydrolysis of salt of strong base and weak acid. Derive the value of Kh for that reaction. Answer: 1. Let us consider the reaction between sodium hydroxide and acetic acid to give sodium acetate and water. NaOH(aq) + CH3COOH(aq) \(\rightleftharpoons\) CH3COONa(aq) + H2O(1)

3. CH3COO is a conjugate base of the weak acid CH3COOH and it has a tendency to react with H + from water to produce unionised acid. But there is no such tendency for Na + to react with OH –

4. CH3COO – (aq) + H2O(1) CH3COOH(aq) + OH – 3 and therefore [OH – ] > [H + ], in such cases, the solution is basic due to the hydrolysis and pH is greater than 7.

Equation (1) x (2) Kh.Ka = [H + ] [OH – ] [H + ] [OH – ] = Kw Kh.Ka = Kw Kh value in terms of degree of hydrolysis (h) and the concentration of salt (c) for the equilibrium can be obtained as in the case of Ostwald’s dilution law Kh = h 2 C and [OH – ] =

COONH

Question 9. Explain about the hydrolysis of salt of strong acid and weak base. Derive Kh and pH for that solution. Answer: 1. Consider a reaction between strong acid HCl and a weak base NH4OH to produce a salt NH4CI and water

2. NH4 is a strong conjugate acid of the weak base NH4OH and it has a tendency to react with OH- from water to produce unionised NH4 as below,

step 3. There is absolutely no such as for instance interest revealed because of the Cl – which [H + ] > [OH – ] the datingranking.net/escort-directory/renton/ solution are acidic while the pH is actually less than seven.

Question 10. Discuss about the hydrolysis of salt of weak acid and weak base and derive pH value for the solution. Answer: 1. Consider the hydrolysis of ammonium acetate CH34(aq) > CH3COO – (aq) + NH + 4(aq)

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