pH of the solution containing 50.0 mL of 0.3 M HCl and 50.0 mL of 0.4 M `NH_(3)` is
[Given `pK_(a)(NH_(4)^(+))=9.26]`

Options

A. 4.74
B. 9.26
C. 8.78
D. 4.63

Answer

C

Explanation

`pK_(a)(NH_(4)^(+))=9.26`
`:. pK_(b)(NH_(3))=pK_(w)-pK_(a)(NH_(4)^(+))`
`=14-9.26=4.74`
On adding 15 m mol (50 mL ) `NH_(3)` the resulting solution contains 15 m mol ( in 100 mL ) of `NH_(4)` Cl and 5 m mol ( in 100 mL ) of `NH_(3)`.
`:. ["Salt"]=15xx10^(-2)M,["Base"]=5xx10^(-2)M`
`:. pOH =pK_(b)+log . (["Salt"])/(["Base"])`
`=4.74+log. (15xx10^(-2))/(5xx10^(-2))`
`=4.74 +log 3=4.74+0.48=5.22`
`pH =14-pOH =14-5.22=8.78`