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 of 0.04 M NaOH is added to the resulting solution ? [Given: log 2 = 0.30 and log 3 = 0.48]

Question

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 of 0.04 M NaOH is added to the resulting solution ? [Given: log 2 = 0.30 and log 3 = 0.48] (A) 5.40 (B) 4.92 (C) None of these (D) 5.88

Tardigrade Admin · Accepted Answer

Total volume =40 mL [ BCl ]=1.6 / 40=0.04 M pH =(1/2)[ pK w - pK b - log C] 5=(1/2)[14- pK b - log 0.04] pK b =5.4

p OH = pK b + log ([ B +]/[ BOH ]) =5.4+ log (1.2/0.4) =5.4+ log 3 =5.4+0.48 =5.88

Q. When a $20 m L$ of $0.08 M$ weak base $BO H$ is titrated with $0.08 H Cl$ , the pH of the solution at the end point is 5. What will be the $pO H$ if $10 m L$ of $0.04 M N a O H$ is added to the resulting solution ? [Given : $lo g 2 = 0.30$ and $lo g 3 = 0.48$ ]

5.40

4.92

None of these

5.88

Total volume $= 40 m L$
$[BCl] = 1.6/40 = 0.04 M$
$p H = \frac{1}{2} [p K_{w} - p K_{b} - lo g C]$
$5 = \frac{1}{2} [14 - p K_{b} - lo g 0.04]$
$p K_{b} = 5.4$

$pO H = p K_{b} + lo g \frac{[ B ^{+} ]}{[ BO H ]}$
$= 5.4 + lo g \frac{1.2}{0.4}$
$= 5.4 + lo g 3$
$= 5.4 + 0.48$
$= 5.88$

Q. When a 20mL of 0.08M weak base BOH is titrated with 0.08HCl, the pH of the solution at the end point is 5. What will be the pOH if 10mL of 0.04MNaOH is added to the resulting solution ? [Given : log2=0.30 and log3=0.48]