Calculate the H+ ion concentration in 0.10 M acetic acid solution. Given that the dissociation constant of acetic acid in water is 1.8 x 10^-5 . Answer is 4.24 x 10^- 4 M

Open in App

Solution

$C H_{3} C O O H \leftrightarrow C H_{3} C O O^{-} + H^{+} T h e d i s s o c i a t i o n c o n s \tan t, K_{a} = \frac{α^{2} C}{1 - α} \sin c e α i s t h e d i s s o c i a t i o n c o n s \tan t . S i n c e t h e a c i d i s a w e a k a c i d, K_{a} = α^{2} C α^{2} = \frac{K_{a}}{C} α = \sqrt{\frac{K_{a}}{C}} = \sqrt{\frac{1.8 \times 10^{- 5}}{0.1}} = 1.3416 \times 10^{- 2} [H^{+}] = α C = 1.3416 \times 10^{- 2} \times 0.1 = 1.3416 \times 10^{- 3} p H = - \log [H^{+}] = - \log [1.3416 \times 10^{- 3}] = 3 \log 10 - \log 1.3416 = 3 - 0.1276 = 2.8724$
Check your answer

Suggest Corrections

Similar questions

Q. The ionization constant of acetic acid is

1.74 \times 10^{- 5}

. Calculate the degree of dissociation of acetic acid in its 0.05 M solution. Calculate the concentration of acetate ion in the solution and its

p H

What would be the hydrogen ion concentration in a solution containing 0.04 mole of acetic acid and 0.05 mole of sodium acetate in 500 ml of the solution?
Dissociation constant for acetic acid is 1.8 x $10^{- 5}$ .
The [ $H^{+}$ ] is:

Calculate the pH of a solution of 0.10 M acetic acid after 50.0 mL of 0.10 M acetic acid solution is treated with 25.0 mL of 0.10 M NaOH. ( $K_{a}$ of acetic acid = $1.8 \times 10^{- 5}$ )

Q. Calculate the degree of ionisation of

01. M

acetic acid. the dissociation constant of acetic acid is

1.8 \times 10^{- 5}

Q. A solution contains

0.4 M

{C H}_{3} C O O H

and

0.2 M

{C H}_{3} C O O N a

. Calculate the concentration of

H^{+}

ions. The ionisation constant of acetic acid is

1.8 \times 10^{- 5}

Join BYJU'S Learning Program