On the salt hydrolysis regarding strong feet and you may weakened acid, we need to get a relationship ranging from K

Question 5. The newest concentration of hydronium ion into the acid boundary solution relies on the latest proportion of concentration of new weak acidic into attention of the conjugate legs within the clear answer. we.age.,

dos. Brand new poor acidic is actually dissociated just to a little extent. Additionally because of prominent ion effect, the fresh dissociation was after that pent up so because of this the harmony intensity of the acid is close to equivalent to the original intensity of the fresh unionised acidic. Likewise the new intensity of the new conjugate base is practically equal to the first intensity of the additional sodium.

step three. [Acid] and [Salt] show the initial intensity of the acid and you may sodium, correspondingly regularly ready yourself new boundary 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 + ,

Get Henderson – Hasselbalch equation Respond to: step 1

5. Furthermore Na is the conjugate acid of your own strong foot NaOH and also zero habit of function having OH

six. This means that there’s no hydrolysis. In such instances [H + ] (OH – ), pH is managed so there fore the answer is actually basic.

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 Lowell MA escort review – ] > [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,

step three. There isn’t any including inclination revealed from the Cl – and that [H + ] > [OH – ] the solution are acid in addition to pH try 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 CH_34(aq) > CH₃COO – _(aq) + NH + ₄(aq)