Calcium lactate is a salt of weak organic acid and strong base represented as `Ca(LaC)_(2).` A saturated solution of `Ca(LaC)_(2)` contains 0.6 mole in 2 litre solution. pOH of solution is 5.60. If `90%` dissociation of the salt takes place then what is `pK_(a)` of lactic acid?
If the pH of a buffer solution containing 0.1 M of monoacidic base and 0.01 M of its salt is 10.5, the pK_(a) of conjugate acid is
The percentage degree of hydrolysis of a salt of weak acid (HA) and weak base (BOH) in its 0.1 M solutions is found to be 10% If the molarity of the solution is 0.05M , the percentage hydrolysis of the salt should be :
Solubility of Ca(OH)_2 in water is 1.5 kgm^(-3) at 25^@ C. Assuming 100% dissociation, calculate the pH of saturated solutions.
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)
In an acidic buffer solution (pH = 4.4) , the concentration ratio of acid and salt is 2 :1 . The value of dissociation constant of weak acid may be
Hydrolysis constant of salt derived from strong acid and weak base is 2xx 10 ^(-5) .The dissociation constant of the weak base is
What happens when a solution of an acid is mixed with a solution of a base in a test tube ? The temperature of the solution increases The temperaure of the solution decreases The temperature of the solution remains constant Salt formation takes place
NARENDRA AWASTHI-IONIC EEQUILIBRIUM-Assertin-Reason Type Questions
