On the sodium hydrolysis out of strong feet and you can weakened acidic, we must get a romance anywhere between K
Question 5. New concentration of hydronium ion in the acidic shield provider utilizes the fresh proportion out of intensity of the fresh weak acid to your concentration of the conjugate legs within the answer. we.elizabeth.,
dos. The fresh weak acid is dissociated just to a small the total amount. Moreover due to common ion perception, the newest dissociation try further pent-up and therefore the brand new harmony intensity of this new acid is close to equal to the first concentration of the fresh unionised acidic. Also brand new concentration of the brand new conjugate base is nearly equal to the first intensity of the added sodium.
3. [Acid] and you will [Salt] represent the first concentration of the new acidic and you will sodium, respectively regularly get ready the fresh new shield service.
Question 6. Explain about the hydrolysis of salt of strong acid Paterson escort service 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) + HNO3(aq) > NaNO3(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 Address: 1
5. Similarly Na is the conjugate acid of solid foot NaOH and has no habit of act that have OH
six. It means there is zero hydrolysis. In these instances [H + ] (OH – ), pH is actually handled and there fore the answer are neutral.
Question 7. Explain about the hydrolysis of salt of strong base and weak acid. Derive the value of Kh 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) Kh.Ka = [H + ] [OH – ] [H + ] [OH – ] = Kw Kh.Ka = Kw Kh 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 Kh = h 2 C and [OH – ] =
COONH
Question 9. Explain about the hydrolysis of salt of strong acid and weak base. Derive Kh 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 is absolutely no such as for example tendency shown because of the Cl – which [H + ] > [OH – ] the solution is acidic and also the pH is actually lower 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 CH34(aq) > CH3COO – (aq) + NH + 4(aq)
