100 mLof $0.01 M {CH}_{3} COOH$ is titrated with $0.01 M NaOH .$ Match the pH of the solution mixture in given column II with the stages of addition of NaOH given in column I. $[{pK}_{a} ({CH}_{3} COOH) = 4.74]$ $(p k_{w} = 14)$
Column I Column II
A 25 mL p 8.22
B 50 mL q 4.26
C 75 mL r 4.74
D 100 mL s 5.22

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

A -q B -r C -s D -p

A -p B -s C -r D -q

A -s B -p C -q D -r

A -s B -p C -r D -q

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} C H_{3} C O O H + N a O H \to C H_{3} C O O H + H_{2} O \\ C H_{3} C O O H = 100 \times 0.01 = 1 m i l li m o l \\ I . 25 m L o f N a O H = 0.25 m i l li m o l \\ I I .50 m L o f N a O H = 0.50 m i l li m o l \\ I I I .75 m L o f N a O H = 0.75 m i l li m o l \\ I V .100 m L o f N a O H = 1.00 m i l li m o l \end{array}$

Stage

$[C H_{3} C O O H]$ Millimol

$[C H_{3} C O O K]$ Millimol

$p H = p K_{a} + \log \frac{[C H_{3} C O O K]}{[C H_{3} C O O H]}$

II.

III.

IV.

0.75

0.50

0.25

0.00

0.25

0.50

0.75

1.00

$4.74 + \log \frac{1}{3} = 4.26$

4.74

$4.74 + \log 3 = 5.22$

8.22

$\begin{array}{l} C H_{3} C O O^{-} + H_{2} O \to C H_{3} C O O H + O H^{-} \\ p H = 7 + \frac{p K_{a}}{2} + \frac{\log C}{2} \end{array}$

Also total volume is $= 200 m L$ containing 1 millimol of $C H_{3} C O O K$
$\begin{array}{l} [C H_{3} C O O K] = \frac{1 \times 10^{- 3} m o l}{0.2 L} = 0.005 M \\ p H = 7 + \frac{4.74}{2} + \frac{\log 0.005}{2} \\ = 7 + 2.37 - 1.15 = 8.22 \end{array}$
$(i) \to (q); (i i) \to (r); (i i i) \to (s); (i v) \to (p)$