Regarding the salt hydrolysis regarding strong foot and you may poor acid, we should instead get a romance ranging from K

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Matter 5. The brand new concentration of hydronium ion inside acid boundary services utilizes the brand new proportion from intensity of the brand new weak acid towards the concentration of its conjugate foot present in the solution. we.age.,

dos. Brand new weakened acidic was dissociated just to a tiny the amount. Additionally due to prominent ion perception, the newest dissociation are subsequent stored so because of this the fresh equilibrium intensity of this new acid is practically comparable to the first concentration of the latest unionised acid. Furthermore the new concentration of the newest conjugate feet is nearly equivalent to the initial concentration of the added salt.

step three. [Acid] and [Salt] represent the initial concentration https://datingranking.net/escort-directory/henderson/ of the brand new acidic and you will salt, correspondingly accustomed ready yourself new barrier provider.

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 + ,

Get Henderson – Hasselbalch formula Answer: step one

5. Likewise Na is the conjugate acidic of one’s good ft NaOH and has now zero habit of respond having OH

six. This means that there surely is zero hydrolysis. In such cases [H + ] (OH – ), pH is actually maintained there fore the clear answer try natural.

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,

step 3. There’s absolutely no like inclination shown by the Cl – and therefore [H + ] > [OH – ] the answer is actually acid additionally the pH are 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 CH_34(aq) > CH₃COO – _(aq) + NH + ₄(aq)