1.

pH of 0.01 M (NH_4)_2 SO_4 and 0.02 M NH_4OH buffer (PK_a of NH_(4)^(+)= 9.26) is

Answer»

`4.74 + log2`
`4.74 - log 2`
`4.74 + LOG1`
`9.26+ log 1`

Solution :Itsisan example of basicbufferfor that
` PH =pK_a+log(["Base"])/(["salt"])`
`=9.26 + log(0.02 )/( 2 xx 0.001 )= 9.26+Log1`


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