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# From the salt hydrolysis out-of good feet and you will weak acid, we have to derive a love between K

From the salt hydrolysis out-of good feet and you will weak acid, we have to derive a love between K

Question 5. The fresh intensity of hydronium ion inside the acidic barrier solution hinges on the proportion regarding intensity of this new poor acid to your concentration of its conjugate foot within the clear answer. i.age.,

dos. This new poor acid are dissociated just to a little the quantity. More over on account of popular ion effect, the newest dissociation try then pent up and hence the newest equilibrium intensity of the newest acidic is virtually comparable to the initial intensity of the new unionised acid. Likewise this new concentration of this new conjugate foot is virtually equivalent to the initial concentration of the additional salt.

step step three. [Acid] and you may [Salt] represent the first intensity of this new acidic and you may sodium, correspondingly accustomed prepare the fresh new 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) + 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 + ,

## Derive Henderson – Hasselbalch picture Respond to: step 1

5. Similarly Na ‘s the conjugate acid of one’s solid ft NaOH and contains zero habit of react having OH

6. This means that there is zero hydrolysis. In these instances [H + ] (OH – ), pH was managed there fore the answer is basic.

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 https://datingranking.net/escort-directory/sandy-springs/ 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 3. There is no such as for example inclination revealed from the Cl – hence [H + ] > [OH – ] the answer was acid and also the pH try below 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)