On the sodium hydrolysis off good ft and you can weak acid, we need to derive a love between K

Concern 5. The fresh concentration of hydronium ion in the acid boundary service depends on the latest proportion out of concentration of the latest poor acidic on amount of their conjugate legs found in the clear answer. we.elizabeth.,

dos. New weak acidic is actually dissociated in order to a tiny the amount. Furthermore on account of well-known ion impact, the fresh new dissociation is actually subsequent suppressed so because of this new balance concentration of the fresh new acid is virtually equivalent to the first concentration of the brand new unionised acidic. Also the fresh concentration of this new conjugate feet is close to equal to the initial concentration of the additional sodium.

step three. [Acid] and you can [Salt] depict the first concentration of brand new acidic and you may sodium, respectively used to prepare yourself the new shield 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) + 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 picture Answer: step 1

5. Similarly Na ‘s the conjugate acidic of the strong base NaOH features zero habit of operate having OH

six. It means that there is no hydrolysis. In such instances [H + ] (OH – ), pH are was able and there fore the answer are natural.

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 Glendale escort service = 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’s absolutely no particularly inclination revealed by Cl – and that [H + ] > [OH – ] the clear answer are acidic and pH is actually 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 CH34(aq) > CH3COO – (aq) + NH + 4(aq)