Regarding salt hydrolysis off strong foot and you can weak acidic, we need to derive a relationship ranging from K

28 Jun Regarding salt hydrolysis off strong foot and you can weak acidic, we need to derive a relationship ranging from K

Concern 5. New concentration of hydronium ion during the acid shield solution hinges on the new ratio from concentration of the latest poor acidic into amount of their conjugate feet within the answer. we.elizabeth.,

dos. New weakened acidic is dissociated simply to a tiny extent. Also because of preferred ion impression, the new dissociation is actually subsequent stored and hence the brand new harmony concentration of the fresh new acidic is close to equivalent to the original concentration of the fresh unionised acid. Similarly brand new intensity of the fresh new conjugate base is nearly equal to the first concentration of the additional sodium.

step three. [Acid] and [Salt] portray the first concentration of the acidic and you can salt, respectively regularly prepare yourself the fresh barrier 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) + 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 + ,

Derive Henderson – Hasselbalch picture Respond to: 1

5. Likewise Na is the conjugate acidic of one’s good ft NaOH and it has zero habit of perform which have OH

six. This means that there is no hydrolysis. In these instances [H + ] (OH – ), pH is maintained so there fore the solution was simple.

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 https://datingranking.net/escort-directory/richardson/ – ] > [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 isn’t any like interest shown because of the Cl – and therefore [H + ] > [OH – ] the answer is acid and the pH are 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 CH_34(aq) > CH₃COO – _(aq) + NH + ₄(aq)