`pK_(b)` of `NH_(3)` is 4.74. The pH when 100 mL of 0.01 M `NH_(3)` solution is 50% neutralised by 0.01 M HCl is
`pK_(b)` of `NH_(3)` is 4.74. The pH when 100 mL of 0.01 M `NH_(3)` solution is 50% neutralised by 0.01 M HCl is
A. 4.74
B. 2.37
C. 9.26
D. 9.48
6 views
1 Answers
Correct Answer - C
At `50%` neutralisation
`[NH_(4)^(+)]=[NH_(3)]`
The solution is a buffer
`:. pOH =pK_(b)+log.([NH_(4)^(+)])/([NH_(3)])`
`4.74+log 1 =4.74`
`pH =14-4=9.26`
6 views
Answered