Depression in freezing point of 0.01 molal aqueous solution of HA is 0.01924 K. The p Ka value of HA is [. Given = molarity = molality, [ mathrmK mathrmf( mathrmH2 mathrmO)=1.85 mathrmKkg mathrmmol-1, log 2=0.3]

Question

Depression in freezing point of 0.01 molal aqueous solution of HA is 0.01924 K. The p Ka value of HA is [. Given = molarity = molality, [ mathrmK mathrmf( mathrmH2 mathrmO)=1.85 mathrmKkg mathrmmol-1, log 2=0.3] (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Δ Tf=i × m × Kf 0.01924= i × 0.01× 1.85 i =1.04 i =1+α( n -1) 1.04=1+α(2-1) α=0.04 mathrmK mathrma=α2 mathrmc=(0.04)2 × 0.01=16 × 10-6 pK a =4.8

Q. Depression in freezing point of $0.01$ molal aqueous solution of $H A$ is $0.01924 K$ . The $p K_{a}$ value of $H A$ is
$[$ Given $=$ molarity $=$ molality, $[K_{f} (H_{2} O) = 1.85 Kkg mol^{- 1}, lo g 2 = 0.3]$

$Δ T_{f} = i \times m \times K_{f}$
$0.01924 = i \times 0.01 \times 1.85$
$i = 1.04$
$i = 1 + α (n - 1)$
$1.04 = 1 + α (2 - 1)$
$α = 0.04$
$K_{a} = α^{2} c = (0.04)^{2} \times 0.01 = 16 \times 1 0^{- 6}$
$p K_{a} = 4.8$

Q. Depression in freezing point of 0.01 molal aqueous solution of HA is 0.01924K. The pKa​ value of HA is [ Given = molarity = molality, [Kf​(H2​O)=1.85Kkgmol−1,log2=0.3]

ΔTf​=i×m×Kf​ 0.01924=i×0.01×1.85 i=1.04 i=1+α(n−1) 1.04=1+α(2−1) α=0.04 Ka​=α2c=(0.04)2×0.01=16×10−6 pKa​=4.8

Q. Depression in freezing point of $0.01$ molal aqueous solution of $H A$ is $0.01924 K$ . The $p K_{a}$ value of $H A$ is
$[$ Given $=$ molarity $=$ molality, $[K_{f} (H_{2} O) = 1.85 Kkg mol^{- 1}, lo g 2 = 0.3]$

$Δ T_{f} = i \times m \times K_{f}$
$0.01924 = i \times 0.01 \times 1.85$
$i = 1.04$
$i = 1 + α (n - 1)$
$1.04 = 1 + α (2 - 1)$
$α = 0.04$
$K_{a} = α^{2} c = (0.04)^{2} \times 0.01 = 16 \times 1 0^{- 6}$
$p K_{a} = 4.8$