What is the pOH of 0.1 M KB (salt of weak acid and strong base) at `25^(@)C` ? (Given : `pK_(b) of B^(-)`=7)

What is the pH of a salt which was produced by a strong acid and a strong base ?

What are the strengths of conjuate acids of a strong base and weak base?

A solution contains equimolar concentration of a weak acid HA and its conjugate base A^(-) , p K_(b) of A^(-) is 9 . The pH of the solution is

The pH of 0.1 M solution of weak mono protic acid is 4.0. Calculate its [H^(+)] and Ka .

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)

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