Home
Class 12
CHEMISTRY
If pK(b) for the reaction of MeNH(2) in ...

If `pK_(b)` for the reaction of `MeNH_(2)` in water is `3.4` the `pK_(a)` for the conjugate acid `Meoverset(+)NH_(3)` will the .

A

`14//3.4`

B

`14+3.4`

C

`(14)(3.4)`

D

`14-3.4`

Text Solution

Verified by Experts

The correct Answer is:
D

Because
`pK_(a)+pK_(b)=pK_(w)=14 (at 25^(@)C)` .
Promotional Banner

Topper's Solved these Questions

  • AMINES

    R SHARMA|Exercise Follow -up Test 3|1 Videos
  • AMINES

    R SHARMA|Exercise Follow -up Test 4|1 Videos
  • AMINES

    R SHARMA|Exercise Follow -up Test 2|1 Videos
  • ALCOHOL, PHENOL AND ETHERS

    R SHARMA|Exercise ETHERS level IV|7 Videos
  • BIOMOLECULES

    R SHARMA|Exercise Archives|47 Videos

Similar Questions

Explore conceptually related problems

If pK_(a) of a weak acid is 5, then pK_(b) of the conjugate base will

If K_(a)=10^(-5) for a weak acid, then pK_(b) for its conjugate base would be

K_(a) for a weak monobasic acid is 1.0xx10^(-6) . The pK_(b) of its conjugate base base is

The pK_(a) for acid A is greater than pK_(a) for acid B, the strong acid is :-

If K_(a) and K_(b) are the dissociation constants of weak acid and its conjugate base , pK_(a) + pK_(b)

When a salt reacts with water to form acidic or basic solution , the process is called hydrolysis . The pH of salt solution can be calculated using the following relations : pH = 1/2 [pK_(w) +pK_(a) + logc] (for salt of weak acid and strong base .) pH = 1/2 [pK_(w) - pK_(b) - logc] (for salt of weak base and strong acid ) . pH = 1/2 [ pK_(w)+pK_(a)-pK_(b)] (for weak acid and weak base ). where 'c' represents the concentration of salt . When a weak acid or a weak base not completely neutralized by strong base or strong acid respectively , then formation of buffer takes place . The pH of buffer solution can be calculated using the following relation : pH = pK_(a) + log . (["Salt"])/(["Acid"]) , pOH = pK_(b) + log . (["Salt"])/([ "Base"]) Answer the following questions using the following data : pK_(a) = 4.7447 , pK_(b) = 4.7447 ,pK_(w) = 14 When 100 mL of 0.1 M NH_(4)OH is added to 50 mL of 0.1M HCl solution , the pH is

When a salt reacts with water to form acidic or basic solution , the process is called hydrolysis . The pH of salt solution can be calculated using the following relations : pH = 1/2 [pK_(w) +pK_(a) + logc] (for salt of weak acid and strong base .) pH = 1/2 [pK_(w) - pK_(b) - logc] (for salt of weak base and strong acid ) . pH = 1/2 [ pK_(w)+pK_(a)-pK_(b)] (for weak acid and weak base ). where 'c' represents the concentration of salt . When a weak acid or a weak base not completely neutralized by strong base or strong acid respectively , then formation of buffer takes place . The pH of buffer solution can be calculated using the following relation : pH = pK_(a) + log . (["Salt"])/(["Acid"]) , pOH = pK_(b) + log . (["Salt"])/([ "Base"]) Answer the following questions using the following data : pK_(a) = 4.7447 , pK_(b) = 4.7447 ,pK_(w) = 14 0.001 M NH_(4)Cl aqueous solution has pH :