MyQuestionIcon

1

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

Lewis Theory

During the ti...

Question

During the titration of a weak diprotic acid $(H_{2} A)$ against a strong base $(N a O H)$ , the $p H$ of the solution half-way to the first equivalent point and that at the first equivalent point are given respectively by :

A

p K_{a_{1}}

and

p K_{a_{1}} + p K_{a_{2}}

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

B

\sqrt{K_{a_{1}} C}

and

\frac{p K_{a_{1}} + p K_{a_{2}}}{2}

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

C

p K_{a_{1}}

and

\frac{p K_{a_{1}} + p K_{a_{2}}}{2}

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

D

p K_{a_{1}}

and

p K_{a_{2}}

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

Open in App

Solution

The correct option is C $p K_{a_{1}}$ and $\frac{p K_{a_{1}} + p K_{a_{2}}}{2}$
During midway of first equivalent point, buffer is formed and the concentration of salt and acid is 1. So, using the formula of buffer, pH = pKa + 0 = pKa.
During first equivalent point, pH is found by using by using the average values of first ionisation and second constant.
The final equivalence point is attained by adding another $10 m l$ ,or a total of $30 m l$ of the titrant to the weak polyprotice acid. image created by Heather Yee.

Suggest Corrections

thumbs-up

0

Similar questions

Q. Which of the following relation is correct regarding

p H

of salt of amphiprotic ion,

H_{2} P O_{4}^{-}

K_{a_{1}}

is first acid dissociation constant of

H_{3} P O_{4}

K_{a_{2}}

is second acid dissociation constant of

H_{3} P O_{4}

.

K_{a 1}

K_{a 2}

are the ionization constants of

H_{3} N^{+} C H I C O O H

H_{3} N^{+} C H I C O O^{-},

respectively, the pH of the solution at the isoelectric point is

⎛ ⎜ ⎜ ⎝ H O O C - C | H^{\oplus} N H_{3} - C H_{2} C O O H ⎞ ⎟ ⎟ ⎠

The structure of aspartic acid is given.The

p K_{a_{1}}, p K_{a_{2}}

p K_{a_{3}}

(A)

respectively are:

1.88, 3.65

9.60

p K_{a_{1}}

corresponds to the ionisation of the

C O O H

⎛ ⎜ ⎜ ⎜ ⎜ ⎝ R \oplus N H_{3} | C H - C O O H ⎞ ⎟ ⎟ ⎟ ⎟ ⎠

p K_{a_{2}}

corresponds to the ionisation of

(\oplus N H_{3})

(ammonium ion). What is the

p H

at isoelectric points

(p I)

?

pH of some solution is given by

p H = \frac{p K_{a_{1}} + p K_{a_{2}}}{2}

. This formula is valid for the compound

N a H_{2} B O_{3}

.

Q. Calculate approximate

p H

of the resultant solution formed by titration of

25 m L

0.4 M N a_{2} C O_{3}

50 m L

0.025 M H C l

p K_{a_{1}} = 6.4

p K_{a_{2}} = 10.3

H_{2} C O_{3}

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Acids and Bases

CHEMISTRY

Watch in App

explore_more_icon

Explore more

Standard XII Chemistry

Join BYJU'S Learning Program

CrossIcon