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

Van't Hoff Factor

The pH of 0...

Question

The pH of $0.1$ M acetic acid solution is closest to[Dissociation constant of the acid, $K_{a} = 1.8 \times 10^{- 5}$ ]

2.87

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

1.00

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

2.07

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

4.76

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

Open in App

Solution

The correct option is A $2.87$
$C H_{3} C O O H ⇌ C H_{3} C O O^{-} + H^{+}$
$C$
$C - C α$ $C α$ $C α$

$K_{a} = \frac{[C H_{3} C O O^{-}] [H^{+}]}{[C H_{3} C O O H]} = \frac{C α C α}{C (1 - α)} = \frac{C α^{2}}{1 - α} = C α^{2} = 1.8 \times 10^{- 5} (1 - α)$

On solving we get $α = 0.0134$

$[H^{+}] = C α = 0.00134$

$p H = - l o g ([H^{+}])$

$p H = 2.87$

Suggest Corrections

Similar questions

5.0 m L

0.1 M N a O H

50 m L

0.1 M

p H

(K a = 1.8 \times 10^{- 5})

0.1

(K_{a} = 10^{- 5})

p H

0.01 m o l

N a O H

1 L

0.1 M

(C H_{3} C O O H)

0.1 M

(C H_{3} C O O N a)

K_{a}

25^{\circ} C

1.8 \times 10^{- 5}

p H = 7

K_{a}

{C H}_{3} C O O H = 1.8 \times 10^{- 5}

K_{a} = 1.8 \times 10^{- 5}

Join BYJU'S Learning Program