QJEEM22Ch863c: JEE Main 2022 Chemistry - Class 11 JEEM22Ch863c Solution: 50 mL of 0.1 M CH 3 COOH is being titrated against 0.1 M NaOH.…

CH₃COOH + NaOH $\to$ CH₃COONa + H₂O

After adding 25 ml of NaOH volume of mixture = 50 + 25 = 75 ml

Initially,

Number of millimole of NaOH = 25 $\times$ 0.1 = 2.5 mm

Number of millimole of CH₃COOH = 50 $\times$ 0.1 = 5 mm

After nutrilisation,

Millimole of NaOH = 0

Millimole of CH₃COOH = 5 $-$ 2.5 = 2.5 mm

Millimole of CH₃COONa = 2.5

After nutrilisation,

Concentration of CH₃COOH = $[C{H_3}COOH] = {{5 - 2.5} \over {75}} = {1 \over {30}}$

Concentration of CH₃COONa = $[C{H_3}COONa] = {{ 2.5} \over {75}} = {1 \over {30}}$

${P^H} = {P^{Ka}} + \log {{[C{H_3}COONa]} \over {[C{H_3}COOH]}}$

$= 4.76 + \log {{{1 \over {30}}} \over {{1 \over {30}}}}$

$= 4.76 + \log (1)$

$= 4.76 + 0$

$= 4.76$

$= 4.76 \times {10^{ - 2}}$