0.1 M $C H_{3}$ COOH is titrated with 0.1M NaOH solution. What would be the difference in pH between $\frac{1}{4}$ and $\frac{3}{4}$ stages of neutralization of acid

2 l o g \frac{3}{4}

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

l o g \frac{1}{3}

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

2 l o g \frac{1}{4}

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

2 l o g 3

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

Open in App

Solution

The correct option is D $2 l o g 3$
$A t \frac{1}{4} t h s t a g e$
$p H = p K_{a} + l o g \frac{\frac{1}{4}}{\frac{3}{4}} = p k_{a} - l o g 3$ $A t \frac{3}{4} t h s t a g e$
$p H_{2} = p K_{a} + l o g \frac{\frac{3}{4}}{\frac{1}{4}} = p k_{a} + l o g 3$ $∴ Δ p H = 2 l o g 3$

Suggest Corrections

Similar questions

0.1M acetic acid solution is titrated aginst 0.1 M NaoH solution. What would be the difference in pH between 1/4 and 3/4 stages of neutralization of acid?

Q. (a) When 100 ml of 0.1 M NaCN solution is titrated with 0.1 M HCI solution the variation of pH of solution with volume of HCI added will be :
(b) Variation of degree of dissociation

α

with concentration for a weak electrolyte at a particular temperature is best represnted by :
(c) M acetic acid solution is titrated against M NaOH solution. the difference in pH between 1/4 and 3/4 stages of neutralization of acid will be 2 log 3.

Q. A

100 m L

solution of

0.1 M C H_{3} C O O H

is titrated with

0.1 M N a O H

, calculate the

p H

25 %

completion and

75 %

completion of the titration.

(p k_{a} = 4.74)

Q. What is the difference in pH for 1/3 and 2/3 stages of neutralization of 0.1 M

C H_{3} C O O H

with 0.1 M

N a O H

Join BYJU'S Learning Program

0.1 M CH3COOH is titrated with 0.1M NaOH solution. What would be the difference in pH between 14 and 34 stages of neutralization of acid

The correct option is D 2 log 3At 14 th stage pH=pKa+log1434=pka−log 3 At 34th stage pH2=pKa+log3414=pka+log 3 ∴ΔpH=2 log 3

0.1 M $C H_{3}$ COOH is titrated with 0.1M NaOH solution. What would be the difference in pH between $\frac{1}{4}$ and $\frac{3}{4}$ stages of neutralization of acid

The correct option is D $2 l o g 3$
$A t \frac{1}{4} t h s t a g e$
$p H = p K_{a} + l o g \frac{\frac{1}{4}}{\frac{3}{4}} = p k_{a} - l o g 3$ $A t \frac{3}{4} t h s t a g e$
$p H_{2} = p K_{a} + l o g \frac{\frac{3}{4}}{\frac{1}{4}} = p k_{a} + l o g 3$ $∴ Δ p H = 2 l o g 3$