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 )

When a 20 mL of 0.08 M weak base BOH is titrated with 0.08 HCl , the p

1.	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 )
Answer» <P>` 5.40` ` 5.88` ` 4.92` None of these Solution :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`