On salt hydrolysis out-of good legs and you will poor acid, we should instead obtain a love between K

Matter 5. The fresh new intensity of hydronium ion inside acidic shield services hinges on the fresh ratio out of intensity of the fresh poor acid for the attention of their conjugate legs found in the clear answer. we.age.,

dos. The brand new weakened acid is dissociated just to a little extent. Moreover because of well-known ion impression, the newest dissociation try subsequent suppressed thus the brand new harmony intensity of the latest acidic is virtually equivalent to the initial concentration of the fresh new unionised acidic. Similarly this new concentration of the latest conjugate legs is practically comparable to the original intensity of the additional salt.

3. [Acid] and [Salt] depict the original intensity of new acid and you will sodium, respectively accustomed prepare yourself new shield services.

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) + HNO_3(aq) > NaNO_3(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 + ,

Obtain Henderson – Hasselbalch formula Respond to: step one

5. Similarly Na is the conjugate acidic of one’s good ft NaOH and has no habit of operate that have OH

six. It indicates that there surely is no hydrolysis. In such instances [H + ] (OH – ), pH is managed there fore the solution was neutral.

Question 7. Explain about the hydrolysis of salt of strong base and weak acid. Derive the value of K_h 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) K_h.K_a = [H + ] [OH – ] [H + ] [OH – ] = K_w K_h.K_a = K_w K_h 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 K_h = h 2 C and [OH – ] =

COONH

Question 9. Explain about the hydrolysis of salt of strong acid and weak base. Derive K_h 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,

3. There is no for example tendency found by Cl – and this [H + ] > [OH – ] the clear answer is acid in addition to pH is actually below 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 CH_34(aq) > CH₃COO – _(aq) + NH + ₄(aq)