A weak monobasic acid HA has K_a=1.0x10^-4. Use K_a=(Calpha^2)/(1-alpha) (solve the quadratic), molality approx molarity

Question

A weak monobasic acid HA has K_a=1.0x10^-4. Use K_a=(Calpha^2)/(1-alpha) (solve the quadratic), molality approx molarity, i=1+alpha and K_f=1.86 K kg mol^-1. Compare the 0.0800 and 0.0200 mol kg^-1 solutions. Which statements are correct?

Accepted Answer

The degree of dissociation is larger in the 0.0200 solution than in the 0.0800 solution

Accepted Answer

The ratio Delta T_f/(K_fm) is closer to 1 for the 0.0800 solution than for the 0.0200 solution

Accepted Answer

In the 0.0800 solution, alphaapprox0.035

Answer

The depression of freezing point of the 0.0200 solution is exactly one-quarter of that of the 0.0800 solution

A weak monobasic acid HA has $K_{a} = 1.0 \times 1 0^{- 4}$ . Use $K_{a} = \frac{C α ^{2}}{1 - α}$ (solve the quadratic), molality $\approx$ molarity, $i = 1 + α$ and $K_{f} = 1.86 K kg mo l^{- 1}$ . Compare the $0.0800$ and $0.0200 mol k g^{- 1}$ solutions. Which statements are correct?

Answer: A, B, C

More questions on van't Hoff factor: dissociation and association

A weak monobasic acid HA has Ka​=1.0×10−4. Use Ka​=1−αCα2​ (solve the quadratic), molality ≈ molarity, i=1+α and Kf​=1.86 K kg mol−1. Compare the 0.0800 and 0.0200 mol kg−1 solutions. Which statements are correct?

Answer: A, B, C

More questions on van't Hoff factor: dissociation and association

A weak monobasic acid HA has $K_{a} = 1.0 \times 1 0^{- 4}$ . Use $K_{a} = \frac{C α ^{2}}{1 - α}$ (solve the quadratic), molality $\approx$ molarity, $i = 1 + α$ and $K_{f} = 1.86 K kg mo l^{- 1}$ . Compare the $0.0800$ and $0.0200 mol k g^{- 1}$ solutions. Which statements are correct?