A solution of 0.1 mole of `CH_3 NH_2 ( K_b =5 xx 10^(-4)) ` and 0.08 mole of HCl is diluted to one litre, then the pOH of the solution is (log 1.25 =0.1)

Options

A. 10.1
B. 3.9
C. 4.9
D. 9.9

Answer

B

Explanation

Given,
`k_b" for "CH_3NH_2=5xx10^(-4)`
`"Moles of "CH_3NH_2=0.1 mol`
`"Moles of "HCl=0.08 mol`
`:'" "pOH=pK_b+log([base])/([acid])`
`=pK_b+log0.1/0.08`
`:.pOH=pK_b+log(1.25)" and "pK_b=-logK_b`
`=-log(5xx10^(-4))`
`=-log5+4log10=-0.69+4`
`pK_b=3.31`
Thus, `pOH=pK_b+log(1.25)`
`=3.31+0.1=3.41`