When a 20 mL of 0.08 M weak base BOH is titrated with 0.08 M HCl, the pH of the solution at the end point is 5. What will be the pOH if 10 mL of 0.04 M NaOH is added to the resulting solution?
`[Given : log 2=0.30 and log 3=0.48]`

Options

A. `5.40`
B. `5.88`
C. `4.92`
D. `6.2`

Answer

B

Explanation

Hint: `BOH + HCl rarr HCl + H_(2)O`
at equivalence point `N_(1)V_(1) = N_(2)V_(2), V_(2) = 20 ml`
`[BCl] = (20 xx 0.08)/(20 + 20) = 0.04`
`pH = 1/2[pK_(w) - pK_(b) - "log" (C)]`
`pK_(b) = 5.4`
`:. B^(+) + OH^(-) rarr BOH`,
`{:("Basic buffer is formed",,,),("Initial milli moles",1.6,0.4,),("Final milli moles",1.2,"_",0.4):}`
`pOH = pK_(b) + "log" ([B^(+)])/([BOH])`
`= 5.4 + "log" ((1.2)/(0.4)) = 5.4 + 0.48`
`pOH = 5.88`