Throughout the sodium hydrolysis away from strong legs and you may weak acidic, we must obtain a relationship between K

Matter 5. The brand new intensity of hydronium ion into the acidic boundary services utilizes the fresh new ratio out-of concentration of this new poor acid toward attention of the conjugate legs within the solution. i.e.,

dos. The brand new weak acid is dissociated only to a tiny the quantity. Also on account of well-known ion feeling, the newest dissociation is after that suppressed thus the fresh new equilibrium concentration of the newest acid is nearly equivalent to the original concentration of the fresh unionised acid. Furthermore the latest intensity of brand new conjugate base is almost equivalent to the initial concentration of the additional salt.

step three. [Acid] and you will [Salt] depict the initial concentration of the newest acidic and you may sodium, correspondingly familiar with prepare yourself the brand 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) + 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 picture Answer: 1

5. Likewise Na ‘s the conjugate acidic of the strong legs NaOH and contains zero tendency to perform that have OH

6. It means that there is no hydrolysis. In such cases [H + ] (OH – ), pH try managed there fore the clear answer 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 like tendency shown of the Cl – and therefore [H + ] > [OH – ] the answer are acid together with pH was lower than 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 CH_34(aq) > CH₃COO – _(aq) + NH + ₄(aq)