MCQ
$50 \,mL$ of $0.1\, M \,CH _{3} COOH$ is being titrated against $0.1 \,M\, NaOH$. When $25\, mL$ of $NaOH$ has been added, the $pH$ of the solution will be $....\,\times 10^{-2}$. (Nearest integer)

(Given : $\left.pK _{ a }\left( CH _{3} COOH \right)=4.76\right)$

$\log 2=0.30$  $\log 3=0.48$  $\log 5=0.69$  $\log 7=0.84$  $\log 11=1.04$

  • A
    $963$
  • B
    $123$
  • $476$
  • D
    $596$

Answer

Correct option: C.
$476$
c
Moles of $CH _{3} COOH =5\, m\, mole\, moles$ of $NaOH =2.5\, m\, mole$

$NaOH + CH _{3} COOH \longrightarrow CH _{3} COONa + H _{2} O$

$2.5\, m\, mole \quad\quad 2.5 \,m\,mole$

$0 \quad\quad\quad\quad\quad 2.5 m \text { mole } \quad\quad2.5 m \text { mole }$

so buffer is formed

$pH = pKa +\log \left(\frac{2.5 / 75}{2.5 / 75}\right)= pKa$

$pH =4.76$

$=476 \times 10^{-2}$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free