" Inas "pH=9.25." Then "pK_(b)" of "NH_(4)OH

Solution of 0.1 N NH_4OH and 0.1 NH_4Cl has has pH 9.25 Then pK_b of NH_4OH is

Solution of 0.1 N NH_(4)OH and 0.1 N NH_(4)Cl has pH 9.25 , then find out pK_(b) of NH_(4)OH .

Solution of 0.1 N NH_(4)OH and 0.1 N NH_(4)Cl has pH 9.25 , then find out K_(b) of NH_(4)OH .

Solution of 0.1 N NH_4 OH and 0.1 N NH_4CI has pH 9.25 . Then find out pK_b "of" NH_4OH

Amount of (NH_(4))_(2)SO_(4) which must be added to 50mL of 0.2 M NH_(4)OH solution to yield a solution of pH 9.26 is ( pK_(b) of NH_(4)OH=4.74 )

When 20 ml of 0.1 MNH_(4)OH are added to 20 ml of MNH_(4)Cl solution, the pH of the buffer formed is 8.2. What is the pK_(b) of NH_(4)OH ?

What will be the amount of (NH_(4))_(2)SO_(4) (in g) which must be added to 500 mL of 0.2 M NH_(4)OH to yield a solution of pH 9.35? ["Given," pK_(a)"of "NH_(4)^(+)=9.26,pK_(b)NH_(4)OH=14-pK_(a)(NH_(4)^(+))]

What will be the amount of (NH_(4))_(2)SO_(4) (in g) which must be added to 500 mL of 0.2 M NH_(4)OH to yield a solution of pH 9.35? ["Given," pK_(a)"of "NH_(4)^(+)=9.26,pK_(b)NH_(4)OH=14-pK_(a)(NH_(4)^(+))]

The potential the cell at 25^(@)C is Given pK_(b) of NH_(4)OH=4.74