In the sodium hydrolysis of good ft and you will poor acidic, we should instead get a love anywhere between K

Question 5. The fresh new intensity of hydronium ion for the acidic buffer services utilizes the new proportion out of intensity of the new weakened acid towards the amount of their conjugate legs found in the solution. i.elizabeth.,

dos. The latest weakened acid are dissociated only to a small the quantity. Moreover due to preferred ion feeling, the fresh new dissociation is actually then suppressed and therefore new balance concentration of the latest acid is almost equal to the initial concentration of the latest unionised acid. Similarly brand new concentration of the fresh new conjugate feet is close to equivalent to the first intensity of the added salt.

step three. [Acid] and you will [Salt] portray the first concentration of the fresh acidic and you can sodium, respectively used to get ready brand 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 + ,

Derive Henderson – Hasselbalch formula Respond to: step one

5. Furthermore Na ’s the conjugate acid of solid legs NaOH and it has no habit of act which have OH

6. It indicates that there’s zero hydrolysis. In such instances [H + ] (OH – ), pH try was able and there fore the answer is 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 – ] > [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’s no like desire revealed by Cl – and therefore [H + ] > [OH – ] the clear answer is acid in addition escort girl Glendale to pH is less than eight.

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)