[" To prepare a buffer of "pH8.26," amount of "(NH_(4))_(2)SO_(4)" to be "],[" added into "500mL" of "0.01MNH_(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:

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 :

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:

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 )

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)^(+))]