Question

`pH` of solution made by mixing `50mL` of `0.2M NH_(4)CI` and `75mL` of `0.1M NaOH is[pK_(b) of `NH_(3)(aq) = 4.74. log 3 = 0.47]`
A. `7.02`
B. `13.0`
C. `7.02`
D. `9.73`

Answer 1

Correct Answer - D
`{:(,NH_(4)+,NaOHrarr,NH_(4)OH+,NaCI),("mmol",50xx0.2,75xx0.1,,),(,=10,=7.5,,),("mmol",10-7.5,-,7.5,),(,=2.5,,,):}`
`rArr` This will result in a basic buffer.
`rArr pOH = pK_(b) + "log" (["Salt"])/(["Base"])`
`=4.74 +"log" (2.5)/(7.5) = 4.27`
`rArr pH = 14 - 4.27 = 9.73`