The dissociation constant of a weak acid HA is 4.9 × 10 - 8 After making the necessary approximations, calculate (i) pH (ii) OH - concentration in a decimolar solution of the acid. (Water has a pH of 7).

Question

The dissociation constant of a weak acid HA is 4.9 × 10 - 8 After making the necessary approximations, calculate (i) pH (ii) OH - concentration in a decimolar solution of the acid. (Water has a pH of 7). (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Molarity (C) = 0.10, [ H+ ] = √ Ka ⋅ C = 7 × 10 - 5 M (α is negligible) p H = 4.15 [ OH- ] = ( Kw / [ H+ ] ) = ( 10 - 14 / 7 × 10 - 5 ) = 1.43 × 10 - 10 M

Q. The dissociation constant of a weak acid $H A$ is $4.9 \times 1 0^{- 8}$ After making the necessary approximations, calculate
(i) pH
(ii) $O H^{-}$ concentration in a decimolar solution of the acid. (Water has a pH of $7$ ).

Molarity $(C) = 0.10$ ,
$[H^{+}] = K_{a} \cdot C = 7 \times 1 0^{- 5} M$ ( $α$ is negligible)
$p H = 4.15$
$[O H^{-}] = \frac{K _{w}}{[ H ^{+} ]} = \frac{1 0 ^{- 14}}{7 \times 1 0 ^{- 5}} = 1.43 \times 1 0^{- 10} M$

Q. The dissociation constant of a weak acid HA is 4.9×10−8 After making the necessary approximations, calculate (i) pH (ii) OH− concentration in a decimolar solution of the acid. (Water has a pH of 7).

Molarity (C)=0.10, [H+]=Ka​⋅C​=7×10−5M (α is negligible) pH=4.15 [OH−]=[H+]Kw​​=7×10−510−14​=1.43×10−10M

Q. The dissociation constant of a weak acid $H A$ is $4.9 \times 1 0^{- 8}$ After making the necessary approximations, calculate
(i) pH
(ii) $O H^{-}$ concentration in a decimolar solution of the acid. (Water has a pH of $7$ ).

Molarity $(C) = 0.10$ ,
$[H^{+}] = K_{a} \cdot C = 7 \times 1 0^{- 5} M$ ( $α$ is negligible)
$p H = 4.15$
$[O H^{-}] = \frac{K _{w}}{[ H ^{+} ]} = \frac{1 0 ^{- 14}}{7 \times 1 0 ^{- 5}} = 1.43 \times 1 0^{- 10} M$