`pH` of a mixture containing `0.10 M X^(-)` and `0.20 M HX` is: `[pK_(b)(X^(-))=4]`

`pOH=pK_(b)+log((["Salt"])/(["Base"]))`
Let a mol litre^(-1) be concentration of salt., then concentration of base `=(0.29-a)mol//L`.
`4.4=-log1.8xx10^(-5)+log((a)/((0.29-a)))`
`:.a=0.09`
`["Salt"]=0.9M`
& `["Base"]=0.29-0.09=0.20M`

`pH` of a mixture containing `0.10 M X^(-)` and `0.20 M HX` is: `[pK_(b)(X^(-))=4]`

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