Home
Class 11
CHEMISTRY
If a salt of strong acid and weak base h...

If a salt of strong acid and weak base hydrolyses appreciably `(C=0.1),` which of the following formula is to be used to calculate degree of hydrolsis 'alpha'?

A

`alpha=sqrt(K_(w))/(K_(a).a)`

B

`alpha=sqrt(K_(w))/(K_(b).a)`

C

`alpha=sqrt(K_(w))/(K_(a).K_(b)`

D

None of these

Text Solution

Verified by Experts

The correct Answer is:
b
Promotional Banner

Topper's Solved these Questions

  • IONIC EEQUILIBRIUM

    NARENDRA AWASTHI|Exercise Level- 1|1 Videos
  • IONIC EEQUILIBRIUM

    NARENDRA AWASTHI|Exercise Level- 2|11 Videos
  • GASEOUS STATE

    NARENDRA AWASTHI|Exercise Level 3 - Subjective Problems|1 Videos
  • SOLID STATE

    NARENDRA AWASTHI|Exercise Level 3 - Subjective Problems|1 Videos

Similar Questions

Explore conceptually related problems

Which of the following not used to test acid or base?

1.0 M solution of which of the following salts is most basic ?

An aqueous solution of the salt is acidic which of the following acids and bases react to give this salt ?

A basic buffer solution contains a weak base B and its conjugate acid BH^(+) . On adding some HCI. which of the following reactions takes place to maintain constant Pt ?

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 :

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

Hydrolysis constant of salt derived from strong acid and weak base is 2xx 10 ^(-5) .The dissociation constant of the weak base is

In the reaction between strong acid and strong base, 1.8 g of water is formed. The heat evolved is used to raise the temperature of one kg of water at 25^@C . What is the final temperature of water?