The `K_(b)` of weak base is `10^(-4)`. The `["salt"]` to `["base"]` ratio to be maintained to keep the `P^(H)` of buffer solution as `9` is .

A weak acid HA has K_(a) = 10^(-6) . What would be the molar ratio of this acid and its salt with strong base so that pH of the buffer solution is 5 ?

A weak acid HA has K_(a) = 10^(-6) . What would be the molar ratio of this acid and its salt with strong base so that pH of the buffer solution is 5 ?

In a buffer solution consisting of a mixture of weak base and its salt, the ratio of salt to base is increases 10 times , the pOH of the solution will

In a buffer solution consisting of a mixture of weak base and its salt, the ratio of salt to base is increases 10 times , the pOH of the solution will

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

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)

The P^(K_(a)) , weak acid is 4.8 what is the ratio of salt to acid, if P^(H) of buffer is 5.8 is to be prepared

The sum of pH and pK_(b) for a basic buffer solution is 13. The ratio of the concentration of the base to that of the salt is

A basic buffer contains equal concentration of base and its salt. The dissociation constant of base is 10^(-6) . Then the pH of the buffer solution is