When a salt reacts with water to form ac...

When a salt reacts with water to form acidic (or ) basic solution the process is called salt hydrolysis. The `P^(H) ` of salt solution can be calculated using the following relation `P^(H) =(1)/(2) [P^(k_w) +P^(Ka) + log C ]` for salt of weak acid and strong base.
`P^(H) =(1)/(2) [P^(K_w) -P^(K_a) -log C ]` for salt of weak base and strong acid ,` P^(H) =(1)/(2) [P^(K_w) +P^(K_a) - P^(K_b) ]`
For a salt of weak acid and weak base where .c. represents the concentration of salt . When a weak acid (or) a weak base is not completly neatralised by strong base (or ) strong acid respectively, then formation of buffer takes places. The `P^(H) ` of buffer solution can be calculated using the following relation
` P^(H) =P^(Ka) +log "" ((["salt")])/(["Acid "]) , P^(OH) =P^(Kb) +log ""(["salt"])/(["base"]) `
Answer the following questions using the following data
`pK_a (CH_3COOH) =4.7447 , pK_b (NH_4OH) =4.7447 , P^(K_W) =14.`
0.001 M `NH_4 Cl` aqueous solution has `P^(H) `

`6.127`

` 7.126 `

` 2.167`

` 1.267`

Verified by Experts

The correct Answer is:

` pH =7 -( pKb+log c)/( 2) =7- ((4.7447-3)/(2))`
` = 6.127`

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When a salt reacts with water to form acidic (or ) basic solution the process is called salt hydrolysis. The P^(H) of salt solution can be calculated using the following relation P^(H) =(1)/(2) [P^(k_w) +P^(Ka) + log C ] for salt of weak acid and strong base. P^(H) =(1)/(2) [P^(K_w) -P^(K_a) -log C ] for salt of weak base and strong acid , P^(H) =(1)/(2) [P^(K_w) +P^(K_a) - P^(K_b) ] For a salt of weak acid and weak base where .c. represents the concentration of salt . When a weak acid (or) a weak base is not completly neatralised by strong base (or ) strong acid respectively, then formation of buffer takes places. The P^(H) of buffer solution can be calculated using the following relation P^(H) =P^(Ka) +log "" ((["salt")])/(["Acid "]) , P^(OH) =P^(Kb) +log ""(["salt"])/(["base"]) Answer the following questions using the following data pK_a (CH_3COOH) =4.7447 , pK_b (NH_4OH) =4.7447 , P^(K_W) =14. One mole CH_3 COOH and one mole CH_3 COONa are dissolved in water one litre aqueous solution The P^(H) of the resulting solution will be

` 9.2553`

` 4.7447 `

` 14`

` 7`

Temperature

Nature of the solution

Degree of dissociation of acid (or) base

Volume of the solution

`3`

`100`

` 1000`

` 15`