`pH` of an aqueous solution of `0.6M NH_(3)` and `0.4M NH_(4)CI` is `9.4 (pK_(b) = 4.74)`. The new `pH` when `0.1M Ca(OH)_(2)` solution is added to it.

Options

A. `9.86`
B. `10.14`
C. `10.2`
D. `10.86`

Answer

A

Explanation

Since `Ca(OH)_(2)` is completely ionised.
`{:(,Ca(OH)_(2)rarr,Ca^(2+)+,2overset(Theta)OH,),("Initla",0.1,0,0,),("Final",c,0,2xx0.1=0.1M,):}`
BBB Rule: On adding base, to the basic buffer, concentration of base increases and salt decreases.
`:.` New concentration of base and salt are:
`["Base"] = [NH_(3)] = 0.6 xx 0.2 = 0.8M`
`["Salt"] = [NH_(4)CI] = 0.4 - 0.2 = 0.2M`
`:. pOH = pK_(b) + log [("Salt")/("Base")]`
`pOH = 4.74 + log ((0.2)/(0.8))`
`= 4.74 - 2 log2 = 4.74 - 2 xx 0.30 = 4.14`
`pH = 14- 4.14 = 9.86`