10 mL of 0-1 M tribasic acid H3A is titrated with 0-1 M NaOH solution- What is the ratio of -H3A-A3- at 2nd equivalence point- -Given that- Ka1 - 10-3- Ka2 - 10-8- Ka3 - 10-12-

The correct option is C ≃ 10^ 7 At the second equivalence point, the ratio [H_3A]/[A^3 ] is given by the expression: [H_3A]/[A^3 ]≈K_a3× K_a1/K ...

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

Byju's Answer

Standard XII

Mathematics

Property 1

10 mL of 0....

Question

$10 m L$ of $0.1 M$ tribasic acid $H_{3} A$ is titrated with $0.1 M N a O H$ solution. What is the ratio of $\frac{[H_{3} A]}{[A^{3 -}]}$ at $2^{n d}$ equivalence point? [Given that: $K_{a_{1}} = 10^{- 3}, K_{a_{2}} = 10^{- 8}, K_{a_{3}} = 10^{- 12}$ ]

≃ 10^{- 4}

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

≃ 10^{+ 4}

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

≃ 10^{- 7}

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

≃ 10^{+ 6}

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

Open in App

Solution

The correct option is C $≃ 10^{- 7}$
At the second equivalence point, the ratio $\frac{[H_{3} A]}{[A^{3 -}]}$ is given by the expression: $\frac{[H_{3} A]}{[A^{3 -}]} \approx \frac{K_{a 3} \times K_{a 1}}{K_{a 2}} = \frac{1 \times 10^{- 12} \times 1 \times 10^{- 3}}{1 \times 10^{- 8}} = 1 \times 10^{- 7} .$

Suggest Corrections

Similar questions

Q. 10 ml of 0.1 M triprotic acid

H_{3} A

is titrated with

0.1 M N a O H

solution. If the ratio of

\frac{[H_{3} A]}{[A^{3 -}]}

2^{n d}

equivalence point is expressed as

10^{- p}

, then

p

is :

[Given :

K_{1} = 10^{- 4}; K_{2} = 10^{- 8}; K_{3} = 10^{- 12}

for

H_{3} A

]

H_{3} A

is a weak triprotic acid

(K_{a_{1}} = 10^{- 5}, K_{a_{2}} = 10^{- 9}, K_{a_{3}} = 10^{- 13})

.
What is the value of

p X

0.1 M H_{3} A (a q .)

solution, where

p X = - log X

and

X = \frac{[A^{3 -}]}{[H A^{2 -}]}

H_{3} A

is a weak triprotic acid

(K a_{1} = 10^{- 5}, K a_{2} = 10^{- 9}, K a_{3} = 10^{- 13}) .

What is the value of

p X

0.1 M H_{3} A (a q)

solution?
Given

p X = - l o g X

and

X = \frac{[A^{3 -}]}{[H A^{2 -}]}

For

M K_{3} A

(potassium salt of a tribasic acid

H_{3} A

) solution:

(Dissociation constants of acid are

k_{a_{1}} k_{a_{2}} & k_{a_{3}}; h << 1

)

25 m L

0.1 N H_{3} P O_{4}

is titrated against

0.1 N N a O H

. The

p H

of the solution after addition of

62.5 m L

of the base is :
(Given :

K_{a 1} = 1 \times 10^{- 3}, K_{a 2} = 1 \times 10^{- 8} a n d K_{a 3} = 1 \times 10^{- 13}

)

Join BYJU'S Learning Program

10 mL of 0.1 M tribasic acid H3A is titrated with 0.1 M NaOH solution. What is the ratio of [H3A][A3−] at 2nd equivalence point? [Given that: Ka1=10−3, Ka2=10−8, Ka3=10−12]

The correct option is C ≃10−7At the second equivalence point, the ratio [H3A][A3−] is given by the expression: [H3A][A3−]≈Ka3×Ka1Ka2=1×10−12×1×10−31×10−8=1×10−7.

$10 m L$ of $0.1 M$ tribasic acid $H_{3} A$ is titrated with $0.1 M N a O H$ solution. What is the ratio of $\frac{[H_{3} A]}{[A^{3 -}]}$ at $2^{n d}$ equivalence point? [Given that: $K_{a_{1}} = 10^{- 3}, K_{a_{2}} = 10^{- 8}, K_{a_{3}} = 10^{- 12}$ ]