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. An organic monoprotic acid

[0.1 M]

is titrated against

0.1 M

N a O H

. By how much does the

p H

change between one fourth and three fourth stages of neutralization? If at one-third stage of neutralization, the

p H

4.45

, what is the dissociation constant of the acid? Between what stages of neutralization may the

p H

change by

2

units?

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0.1 M CH3COOH solution is titrated against 0.05 M NaOH solution. The pH for 0.1 M CH3COOH is 3. The pH at 14th stages of neutralization of acid is:

The correct option is A 4.5The reaction is as follows: CH3COOH+NaOH⟶CH3COONa+H2OBefore addition of NaOH: 0.1After addition of NaOH: 0.1×34 0.1×14At 1/4th neutralization of acid,∵ pH=−logKa+log[Conjugate base][Acid]or, pH=−log10−5+log(0.1/4)(0.3/4)or, pH=4.5

$0.1 M$ ${C H}_{3} C O O H$ solution is titrated against $0.05 M$ $N a O H$ solution. The $p H$ for $0.1 M$ ${C H}_{3} C O O H$ is $3$ . The $p H$ at $\frac{1}{4}$ th stages of neutralization of acid is: