When a solution of a weak monoprotic acid is titrated against a strong base,at the half-neutralization point, which of the following is true?

p H < p k_{a}

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

p H > p k_{a}

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

p H = p k_{a}

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

Can not predict

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

Open in App

Solution

The correct option is C $p H = p k_{a}$
At half neutralization point, one half of the acid is present in the form of salt.
Hence, $[A c i d] = [S a l t]$ .
$p H = p K_{a} + l o g \frac{[s a l t]}{[a c i d]} = p K_{a} + l o g \frac{1}{1} = p K_{a}$

Suggest Corrections

Similar questions

Q. A weak monobasic acid is half neutralised by a strong base. If the pH of the solution is

5.4

, its

p K_{a}

is:

Q. Assertion :In a titration of weak acid with strong base, the

p H

at the half equivalence point is

p K_{a}

Reason: At half equivalence point, it will form acidic buffer at it's maximum capacity where [acid] = [salt]

100 mL of 1.0 M solution of a monobasic acid $(p K_{a} = 5)$ is titrated against $C a (O H)_{2}$ solution. At equivalence point the pH of solution is ?

Q. When

10 m L

0.1 M

acetic acid (

p K_{a} = 5.0

) is titrated against

10 m L

0.1 M

ammonia solution

(p K_{b} = 5.0)

, the equivalence point occurs at pH:

Q. A given weak acid (0.01M ) has

p K_{a} = 6.

the pH of the solution is :

Join BYJU'S Learning Program

When a solution of a weak monoprotic acid is titrated against a strong base,at the half-neutralization point, which of the following is true?

The correct option is C pH=pkaAt half neutralization point, one half of the acid is present in the form of salt.Hence, [Acid]=[Salt].pH=pKa+log[salt][acid]=pKa+log11=pKa

The correct option is C $p H = p k_{a}$
At half neutralization point, one half of the acid is present in the form of salt.
Hence, $[A c i d] = [S a l t]$ .
$p H = p K_{a} + l o g \frac{[s a l t]}{[a c i d]} = p K_{a} + l o g \frac{1}{1} = p K_{a}$