MCQ
A solution is prepared by mixing $0.01 mol$ each of $H _2 CO _3, NaHCO _3, Na _2 CO _3$, and $NaOH$ in $100 mL$ of water. $pH$ of the resulting solution is. . . . . . .

[Given : $p K _{ a 1}$ and $p K _{ a 2}$ of $H _2 CO _3$ are $6.37$ and $10.32 $, respectively $; \log 2=0.30$ ]

  • A
    $10.1$
  • B
    $11.2$
  • $10.2$
  • D
    $10.3$

Answer

Correct option: C.
$10.2$
c
$H _2 CO _3+ NaOH \longrightarrow NaHCO _3+ H _2 O$

Milli moles $10$ $10$ -
At end $0$ $0$ $10+10=20$

Final mixture has 20 milli moles $NaHCO _3$ and 10 milli moles $Na _2 CO _3$

$pH = pKa _2+\log \frac{\text { Salt }}{\text { Acid }}$

$pH = pKa _2+\log \left(\frac{10}{20}\right)  {\left[\text { Buffer : } Na _2 CO _3+ NaHCO _3\right]}$

$=10.32-\log 2=10.02$

[Buffer : $Na _2 CO _3+ NaHCO _3$ ]

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