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2 years ago

Given that a for HCN is 6. 2×10^−10 at 25 °C. What is the value of b for cn− at 25 °C?

Chemistry

1 answer:

Kay [80]2 years ago

4 0

If Ka for HCN is 6. 2×10^−10 at 25 °C, then the value of Kb for cn− at 25 °C is 1.6 × 10^(-5).

<h3>What is base dissociation constant? </h3><h3 />

The base dissociation constant (Kb) is defined as the measurement of the ions which base can dissociate or dissolve in the aqueous solution. The greater the value of base dissociation constant greater will be its basicity an strength.

The dissociation reaction of hydrogen cyanide can be given as

HCN --- (H+) + (CN-)

Given,

The value of Ka for HCN is 6.2× 10^(-10)

The correlation between base dissociation constant and acid dissociation constant is

Kw = Ka × Kb

Kw = 10^(-14)

Substituting values of Ka and Kw,

Kb = 10^(-14) /{6.2×10^(-10) }

= 1.6× 10^(-5)

Thus, the value of base dissociation constant at 25°C is 1.6 × 10^(-5).

learn more about base dissociation constant :

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At 85°C, the vapor pressure of A is 566 torr and that of B is 250 torr. Calculate the composition of a mixture of A and B that b

Phantasy [73]

Answer:

Composition of the mixture:

$x_{A} =0.652=65.2$ %

$x_{B} =0.348=34.8$ %

Composition of the vapor mixture:

$y_{A} =0.809=80.9$ %

$y_{B} =0.191=19.1$ %

Explanation:

If the ideal solution model is assumed, and the vapor phase is modeled as an ideal gas, the vapor pressure of a binary mixture with $x_{A}$ and $x_{B}$ molar fractions can be calculated as:

$P_{vap}=x_{A}P_{A}+x_{B}P_{B}$

Where $P_{A}$ and $P_{B}$ are the vapor pressures of the pure compounds. A substance boils when its vapor pressure is equal to the pressure under it is; so it boils when $P_{vap}=P$ . When the pressure is 0.60 atm, the vapor pressure has to be the same if the mixture is boiling, so:

$0.60*760=P_{vap}=x_{A}P_{A}+x_{B}P_{B}\\456=x_{A}P_{A}+(1-x_{A})P_{B}\\456=x_{A}*(P_{A}-P_{B})+P_{B}\\\frac{456-P_{B}}{P_{A}-P_{B}}=x_{A}\\\\\frac{456-250}{566-250}=x_{A}=0.652$

With the same assumptions, the vapor mixture may obey to the equation:

$x_{A}P_{A}=y_{A}P$ , where P is the total pressure and y is the fraction in the vapor phase, so:

$y_{A} =\frac{x_{A}P_{A}}{P}=\frac{0.652*566}{456} =0.809=80.9$ %

The fractions of B can be calculated according to the fact that the sum of the molar fractions is equal to 1.

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