When a 20 mL of 0.08 M weak base BOH is titrated with `0.08 HCl` , the pH of the solution at the end point is 5. What will be the pOH if 10 mL 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. None of these

Answer

B

Explanation

At end point
` N_1V_1 = N_2V_2 , [BCl ]= (20 xx 0.08)/( 20+ 20 ) =0.04`
` 5= (1)/(2) [14 -P^(K_b) -log 4 xx 10^(-2) ]rArr P^(K_b) =5.4 `
` {:(B^(+) +, OH^(-) to, BOH) ,( 1.6 m " mol " , 0.4 m " moles " ,0),( 1.2 m " mol " ,-, 0.4 m " mole" ):}`
` P^(OH) =5.4 +log ""(1.2)/(0.4) =5.88`