Throughout the sodium hydrolysis away from solid foot and you may poor acidic, we must obtain a romance between K

Throughout the sodium hydrolysis away from solid foot and you may poor acidic, we must obtain a romance between K

Concern 5. This new concentration of hydronium ion inside acidic buffer service depends on the brand new ratio from concentration of the brand new weak acid toward concentration of the conjugate base present in the clear answer. i.e.,

dos. The newest poor acidic was dissociated simply to a tiny the quantity. Additionally due to popular ion perception, the fresh new dissociation try next suppressed thus the latest balance intensity of this new acidic is close to equivalent to the initial concentration of this new unionised acidic. Likewise the newest concentration of the conjugate ft is practically comparable to the original concentration of the added salt.

step 3. [Acid] and you can [Salt] portray the initial concentration of the new acidic and you will salt, correspondingly familiar with ready yourself this new shield solution.

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 formula Respond to: 1

5. Likewise Na is the conjugate acid of the good foot NaOH and it has zero tendency to react that have OH

six. This means that there’s zero hydrolysis. In such cases [H + ] (OH – ), pH are managed so there fore the clear answer is 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 check this site + ], 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 three. There isn’t any instance tendency revealed of the Cl – and this [H + ] > [OH – ] the clear answer are acid and the pH try below 7.

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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