The composition of an acidic buffer mixture made up of HA and NaA of total molarity 0.29 having pH =4.4 and Ka=1.8 × 10-5 in terms of concentration of salt and acid respectively is

Question

The composition of an acidic buffer mixture made up of HA and NaA of total molarity 0.29 having pH =4.4 and Ka=1.8 × 10-5 in terms of concentration of salt and acid respectively is (A) 0.09 M and 0.20 M (B) 0.20 M and 0.09 M (C) 0.1 M and 0.19 M (D) 0.19 M and 0.10 M

Tardigrade Admin · Accepted Answer

pH =- log Ka+ log ([ text Salt ]/[ text Acid ]) Let a mol L-1 be concentration of salt, then concentration of acid =(0.29-a) 4.4=- log (1.8 × 10-5)+ log (a/(0.29-a)) ⇒ a=0.09 ∴ [ Salt ]=0.09 M ; [ Acid ]=0.29-0.09=0.20 M

Q. The composition of an acidic buffer mixture made up of $H A$ and $N a A$ of total molarity $0.29$ having $p H = 4.4$ and $K_{a} = 1.8 \times 1 0^{- 5}$ in terms of concentration of salt and acid respectively is

$0.09 M$ and $0.20 M$

$0.20 M$ and $0.09 M$

$0.1 M$ and $0.19 M$

$0.19 M$ and $0.10 M$

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

Let $a m o l L^{- 1}$ be concentration of salt, then concentration of acid $= (0.29 - a)$

$4.4 = - lo g (1.8 \times 1 0^{- 5}) + lo g \frac{a}{( 0.29 - a )}$

$\Rightarrow a = 0.09$

$∴ [S a lt] = 0.09 M$ ; [ Acid $] = 0.29 - 0.09 = 0.20 M$

Q. The composition of an acidic buffer mixture made up of HA and NaA of total molarity 0.29 having pH=4.4 and Ka​=1.8×10−5 in terms of concentration of salt and acid respectively is

0.09M and 0.20M

0.20M and 0.09M

0.1M and 0.19M

0.19M and 0.10M