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

Byju's Answer

Standard XII

Chemistry

Van't Hoff Factor

Calculate the...

Question

Calculate the hydronium ion concentration and $p H$ at the equivalence point in the reaction of $22.0 m l$ of $0.10 M$ acetic acid, ${C H}_{3} C O O H$ , with $22.0 m L$ of $0.10 M$ $N a O H$ .
$K_{a} = 2 \times 10^{- 5}$

Open in App

Solution

$22 m l$ of $0.1 N$ acetic acid will just neutralise $22 m l$ of $0.1 N$ $N a O H$ . The resulting solution will be neutral.
$p H = 7$
$[H^{+}] = 1 \times 10^{- 7} m o l / L$

Suggest Corrections

Similar questions

Q. Calculate the hydronium ion concentration and pH at the equivalence point in a titration of

50.0 m L

0.40 M {N H}_{3}

with

0.40 M

HCl.

K_{b} = 2 X 10^{- 5}

Q. 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 \times 10^{- 6}

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

{O H}^{-}

concentration at the equivalent point when a solution of

0.1 M

acetic acid is titrated with a solution of

0.1 M

N a O H

K_{a}

for the acid

= 1.9 \times 10^{- 5}

Q. The pH of solution containing

0.10 M

sodium acetate and

0.03 M

acetic acid is: (

p K_{a}

for

{C H}_{3} C O O H = 4.57

)

Join BYJU'S Learning Program

Calculate the hydronium ion concentration and pH at the equivalence point in the reaction of 22.0 ml of 0.10 M acetic acid, CH3COOH, with 22.0 mL of 0.10 M NaOH. Ka =2×10−5

22ml of 0.1N acetic acid will just neutralise 22ml of 0.1N NaOH. The resulting solution will be neutral.pH=7[H+]=1×10−7mol/L

Calculate the hydronium ion concentration and $p H$ at the equivalence point in the reaction of $22.0 m l$ of $0.10 M$ acetic acid, ${C H}_{3} C O O H$ , with $22.0 m L$ of $0.10 M$ $N a O H$ .
$K_{a} = 2 \times 10^{- 5}$

$22 m l$ of $0.1 N$ acetic acid will just neutralise $22 m l$ of $0.1 N$ $N a O H$ . The resulting solution will be neutral.
$p H = 7$
$[H^{+}] = 1 \times 10^{- 7} m o l / L$