In what volume ratio should you mix `1.00 M` solution `NH_(4)Cl` and 1.00 M `NH_(3)` to produce a buffer solution of pH 9.80 ? `[pK_(b)(NH_(3))=4.74]`

Options

A. `1:1.35`
B. `3.5 :1`
C. `2:1`
D. `1:2`

Answer

B

Explanation

`[NH_(4)Cl]=V_(1)mL =V_(1)` millimol
`[NH_(3)]=V_(2)mL =V_(2)` millimol
`=pH =9.80 , pOH =4.2`
`pOH =pK_(a)+log.(V_(1))/(V_(2))`
`4.2 =4.74 +log. (V_(1))/(V_(2))`
`-0.54=log. (V_(1))/(V_(2))`
or `log.(V_(2))/(V_(1))=0.54`
`(V_(2))/(V_(1))=` antilog `0.54=3.5`
Thusl, `(V_(2))/(V_(1))=(3.5 )/(1)`