A weak acid HA after treatment with 1.2 mL of 0.1 M strong base has a pH of 5. At the end point, the volume of same base required is 26.6 mL. The value of textK texta is

Question

A weak acid HA after treatment with 1.2 mL of 0.1 M strong base has a pH of 5. At the end point, the volume of same base required is 26.6 mL. The value of textK texta is (A)  8.2× 10-6  (B)  6.4× 10-6  (C)  5.3 × 10-5  (D)  2.4× 10-6

Tardigrade Admin · Accepted Answer

For complete neutralisation,
Total milliequivalent of acid = milliequivalent of base

= 26.6 \times 0.1 = 2.66

For partial neutralisation;

H A + BO H 2.66 1.46 \to B A + H_{2} O 1.2 0 0 1.2 C - before reaction (C + 1.2) - after reaction.

The resultant mixture has HA and BA and thus acts as a buffer.

∴ p H = - lo g K_{a} + lo g \frac{[ Salt ]}{[ Acid ]}

5 = - lo g K_{a} + lo g \frac{1.2}{1.46} = - lo g K_{a} - 0.085

∴ lo g K_{a} = - 5.085

or

K_{a} =

anti

lo g (- 5.085) = 8.21 \times 1 0^{- 6}

Q. A weak acid HA after treatment with $1.2 m L$ of $0.1 M$ strong base has a pH of 5. At the end point, the volume of same base required is $26.6 m L$ . The value of $K_{a}$ is

$8.2 \times 10^{- 6}$

$6.4 \times 10^{- 6}$

$5.3 \times 10^{- 5}$

$2.4 \times 10^{- 6}$

Q. A weak acid HA after treatment with 1.2mL of 0.1M strong base has a pH of 5. At the end point, the volume of same base required is 26.6mL. The value of Ka​ is

8.2×10−6

6.4×10−6

5.3×10−5

2.4×10−6

Q. A weak acid HA after treatment with $1.2 m L$ of $0.1 M$ strong base has a pH of 5. At the end point, the volume of same base required is $26.6 m L$ . The value of $K_{a}$ is

$8.2 \times 10^{- 6}$

$6.4 \times 10^{- 6}$

$5.3 \times 10^{- 5}$

$2.4 \times 10^{- 6}$