Regan excellentrding the sodium hydrolysis out-of strong foot and you may weak acidic, we must obtain a love ranging from K

Regan <a href="https://datingranking.net/escort-directory/rockford/">https://datingranking.net/escort-directory/rockford/</a> excellentrding the sodium hydrolysis out-of strong foot and you may weak acidic, we must obtain a love ranging from K

Question 5. The fresh new intensity of hydronium ion when you look at the acidic barrier services utilizes the fresh proportion away from concentration of the new poor acid towards attention of their conjugate ft within the answer. i.age.,

dos. The weakened acid is dissociated simply to a tiny extent. Furthermore on account of popular ion impact, the dissociation is actually further suppressed and therefore the newest harmony intensity of the brand new acidic is nearly equal to the original concentration of the latest unionised acid. Similarly the newest intensity of this new conjugate feet is virtually equivalent to the original intensity of the added salt.

3. [Acid] and you will [Salt] show the original concentration of brand new acid and you can salt, respectively used to ready yourself the newest barrier service.

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 + ,

Get Henderson – Hasselbalch formula Answer: 1

5. Similarly Na is the conjugate acidic of your own solid legs NaOH possesses zero tendency to act having OH

6. It means that there surely is zero hydrolysis. In these instances [H + ] (OH – ), pH is actually managed so there fore the solution is actually 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 = 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’s no for example tendency found from the Cl – hence [H + ] > [OH – ] the clear answer is actually acid together with pH try 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 CH34(aq) > CH3COO – (aq) + NH + 4(aq)

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