pH of 0.01 M `(NH_(4))_(2) SO_(4)` and 0.02 M `NH_(4)OH` buffer `(pK_(a)` of `NH_(4)^(+)=9.26)` is

The pH of basic buffer mixtures is given by : pH=pK_(a)+log((["Base"])/(["Salt"])) , whereas pH of acidic buffer mixtures is given by: pH= pK_(a)+log((["Salt"])/(["Acid"])) . Addition of little acid or base although shows no appreciable change for all practical purpose, but since the ratio (["Base"])/(["Salt"]) or (["Salt"])/(["Acid"]) change, a slight decrease or increase in pH results in. The amount of (NH_(4))_(2)SO_(4) to be added to 500 mL of 0.01M NH_(4)OH solution (pK_(a) for NH_(4)^(+) is 9.26) prepare a buffer of pH 8.26 is:

To prepare a buffer of pH 8.26 , amount of (NH_(4))_(2)SO_(4) to be added into 500mL of 0.01M NH_(4)OH solution [pK_(a)(NH_(4)^(+))=9.26] is:

To prepare a buffer of pH 8.26 , amount of (NH_(4))_(2)SO_(4) to be added into 500mL of 0.01M NH_(4)OH solution [pK_(a)(NH_(4)^(+))=9.26] is:

The Ph of basic buffer mixtures is given by : Ph=Pk_(a)+ log (["Base"])/(["Salt"]) whereas Ph of acidic buffer mixtures is given by : Ph = pK_(a)+"log"(["Salt"])/(["Acid"]) . Addition of little acid or base although shows no appreciable change in Ph for all practical purposes, but sicne the ratio (["Base"])/(["Salt"]) or (["Salt"])/(["Acid"]) changes, a slight decrease or increase in pH results. The amount of (NH_(4))_(2)SO_(4) to be added to 500mL of 0.01 M NH_(4)OH solution (pH_(a)NH_(4)^(+) is 9.26) to prepare a buffer of pH 8.26 is :

What is the pH of solution obtained by mixing 100 mL of each 0.2 M NH_(4)Cl and 0.2M NH_(4)OH , if pK_(b) of NH_(4)OH is 4.2 ?

What is the pH of solution obtained by mixing 100 mL of each 0.2 M NH_(4)Cl and 0.2M NH_(4)OH , if pK_(b) of NH_(4)OH is 4.2 ?

The pH of a buffer solution formed by mixing 30ml of 0.1 M NH_(4) OH and 30ml os 1M NH_(4) Cl solution is 8.6. The pK _(b) of NH-(4) OH is

A solution is prepared by mixing 100 ml of 0.1 M NH_(4) OH and 200 ml of 0.2 M NH_(4) Cl . pK_(b) of NH_(4) OH is 4.8 . Then the pH of the solution is