You visited us 1 times! Enjoying our articles? Unlock Full Access!

Salt of Weak Acid and Strong Base

A 50 mL solut...

Question

A $50 m L$ solution of $0.2 M C H_{3} C O O H$ is titrated against $0.2 M N a O H$ . Find the $p H$ of the solution when $50 m L$ of $N a O H$ is added.
$p K_{a} (C H_{3} C O O H) = 4.74$

8.87

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

9.44

No worries! We‘ve got your back. Try BYJU‘S free classes today!

5.66

No worries! We‘ve got your back. Try BYJU‘S free classes today!

10.52

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A 8.87
mmol of $C H_{3} C O O H = 50 \times 0.2 = 10$
mmol of $N a O H = 50 \times 0.2 = 10$

$C H_{3} C O O H (a q) + N a O H (a q) \to C H_{3} C O O N a (a q) + H_{2} O (l) I n i t i a l l y : 101000 F i n a l : 001010$
Since only salt of weak acid and strong base i.e. $C H_{3} C O O N a$ is left, it will undergo hydrolysis.
$Concentration = \frac{Moles}{Total Volume}$

$[C H_{3} C O O N a] = [C H_{3} C O O^{-}] = \frac{10}{100} = 0.1 M$

$C H_{3} C O O^{-} (a q) + H_{2} O (l) ⇌ C H_{3} C O O H (a q) + O H^{-} (a q)$
$p H = 7 + \frac{1}{2} (p K_{a} + l o g C) p H = 7 + \frac{1}{2} (4.74 + l o g (0.1)) p H = 7 + \frac{1}{2} (4.74 - 1) p H = 7 + 1.87 p H = 8.87$
Theory :
Phenolphthalein indicator is used for weak acid and strong base titrations

Suggest Corrections

Similar questions

50 m L

0.2 M C H_{3} C O O H

0.2 M N a O H

p H

0 m L N a O H

p H_{i n i t i a l}

10 m L

N a O H

p H_{10 m L}

K_{a} (C H_{3} C O O H) = 1.8 \times 10^{- 5}

50 m L

0.2 M H C l

0.1 M N a O H

p H

20 m L N a O H

C H_{3} C O O H

N a O H

p H

[p K_{a} (C H_{3} C O O H) = 4.74]

100 m L

0.1 M C H_{3} C O O N a (a q)

0.1 M H C l (a q)

p H

50 m L

H C l

p K_{a} (C H_{3} C O O H) = 4.74 a t 25^{o} C

C H_{3} C O O H

Join BYJU'S Learning Program

Explore more

Join BYJU'S Learning Program